Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1. The diagrams below are graphs of Force in kiloNewtons versus time in milliseconds for the motion of a 5-kg block moving to the right at 4.0 m/s.
   (a) What is the magnitude and direction of the impulse acting on the block in each case?
   (b) What is the magnitude and direction of the average force acting on the block in each case?
   (c) What is the magnitude and direction of the final velocity of the block in each case?
   

   1. Impulse is given by the area under the F-t curves. Since we have simple shapes, it is easy to find the area. For rectangles area is height × base and for triangles area is half the height × base.
      1. I = 3 kN × 3 ms = 9 N-s

      2. I = −1 kN × 6 ms = −6 N-s

      3. I = 2 kN × 2 ms + ½(−4 kN) × 2 ms = 0 N-s

      4. I = ½(4 kN) × 4 ms = 8 N-s

      If the impulse is positive, the net area was above the curve and it is directed to the right, if negative to the left.

       

   2. We know I = F_aveΔt where Δt is how long the collision lasts. We read Δt from the graphs, so F_ave = I/Δt.
      1. F_ave = (9 N-s)/(3 ms) = 3000 N

      2. F_ave = (−6 N-s)/(6 ms) = -1000 N

      3. F_ave = (0 N-s)/(4 ms) = 0 N

      4. F_ave = (8 N-s)/(4 ms) = 2000 N

      If the average force is positive it is directed to the right, if negative to the left. The impulse and force have the same direction.

       

   3. Impulse is also equal to the difference in momentum, I = mv_f − mv_i. We can rearrange our equation for v_f, v_f = I/m + v_i.
      1. v_f = (9 N-s)/(5.0 kg) + 4 m/s = 5.8 m/s

      2. v_f = (?6 N-s)/(5.0 kg) + 4 m/s = 2.8 m/s

      3. v_f = (0 N-s)/(5.0 kg) + 4 m/s = 4 m/s

      4. v_f = (8 N-s)/(5.0 kg) + 4 m/s = 5.6 m/s



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2. The diagrams below are the velocity versus time graphs for the collision of motion of a 4-kg block with a wall. The collision lasts for 20 milliseconds in each case.
   (a) What is the magnitude and direction of the impulse acting on the block in each case?
   (b) What is the magnitude and direction of the average force acting on the block in each case?
   

   1. Impulse is also equal to the difference in momentum, I = mv_f − mv_i. We have the mass, m = 4 kg.
      1. I = (4 kg) × (−6 m/s − 6 m/s) = −48 N-s

      2. I = (4 kg) × (2 m/s − 8 m/s) = −24 N-s

      3. I = (4 kg) × (6 m/s − 0 m/s) = +24 N-s

      If the impulse is positive it is directed to the right, if negative to the left.

   2. We know I = F_aveΔt where Δt is how long the collision lasts. We have already calculated I and we are given Δt = 20 ms, so F_ave = I/Δt.
      1. F_ave = (−48 N-s)/(20 ms) = −2400 N

      2. F_ave = (−24 N-s)/(20 ms) = −1200 N

      3. F_ave = (+24 N-s)/(20 ms) = +1200 N



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3. You've been rowdy and obnoxious in a bar and are now in the process of being thrown out by the scruff of the neck by the bouncer. The bouncer has hold of you for 5.0 s and you are take from a seated position to a final speed of 2.75 m/s. If your mass is 70.0 kg, what was your final momentum? What impulse and average force did the bouncer exert on your person? Assume all motion is in a straight line.

   Momentum is defined by p = mv. Taking the direction of motion as positive, your initial momentum was zero and your final momentum is
   

   p = (70.0 kg)(2.75 m/s) = 192.5 kg-m/s .

   Impulse is defined as the change in momentum
   

   I = p_f - p_i = 192.5 kg-m/s .

   Average force is related to impulse by I = F_averageΔt, so
   

   F_average = I / Δt = 192.5 kg-m/s / 5 s = 38.5 N .

   This is the average force exerted on you and is in the same direction as your motion.
   

   

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4. A ball of mass 0.500 kg with speed 15.0 m/s collides with a wall and bounces back with a speed of 10.5 m/s. If the motion is in a straight line, calculate the initial and final momenta and impulse. If the ball exerted an average force of 1000 N on the wall, how long did the collision last?
   

   Momentum is defined by p = mv. Taking the right as positive, the initial momentum of the ball is
   

   p_i = (0.5 kg)(-15 m/s) = -7.5 kg-m/s .

   The final momentum is
   

   p_f = (0.5 kg)(10.5 m/s) = 5.25 kg-m/s .

   Impulse is defined as the change in momentum
   

   I = p_f - p_i = 12.75 kg-m/s .
   

   Average force is related to impulse by I = F_averageΔt, and the wall would exert this force on the ball to the right. Therefore

   Δt = I / F_average = 12.75 kg-m/s / +1000 = 0.013 s.
   

   The ball is in contact with the wall for approximately 13 milliseconds.
   

   

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5. A ball of mass 0.25 kg glances of a wall as shown in the diagram. The ball approaches at 15 m/s at θ = 30° and leaves at 12 m/s at φ = 20°. The collision lasts for 15 milliseconds.
   (a) What are the components of the impulse experienced by the ball?
   (b) What are the components of the average force acting on the ball?
   
   1. We know Impulse is equal to the difference in momentum, I = mv_f − v_i. This is a vector equation and to get components we consider the x and y components separately.

      I_x = mv_fx − mv_ix = (0.25) × (12cos20° − 15cos30°) = −0.4285 N-s

      I_y = mv_fy − mv_iy = (0.25) × (12sin20° − (−15sin30°)) = +2.9011 N-s

      or I = −i0.4285 + j2.9011 N-s.

   2. We know I = F_aveΔt where Δt is how long the collision lasts. We have already calculated I and we are given Δt = 15 ms, so F_ave = I/Δt

      so F_ave = (−i0.4285 + j2.9011 N-s) / (15 ms) = −i28.6 + j193.4 N.



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6. Explain why a person wearing a seatbelt in a car accident is less likely to be seriously hurt than the person who isn't wearing a seatbelt.

   Wearing a seatbelt would not effect the person's initial and final momentum. Since impulse is the change in momentum, the impulse experienced by the person would be the same in either case. However, it is not impulse which is dangerous but the magnitude of the forces acting on the person's body. Impulse and average force are related by I = F_average Δt. For the same impulse, the average force will be higher as the collision time Δt decreases. Cars are designed to crumple on impact. This crumpling is designed to make a collision last as long as possible. If a person is wearing a seatbelt, the time for the person's change in momentum is the same as that of the car thereby minimizing the force on the occupant. If the person is not wearing a seatbelt, the Law of Inertia dictates that the person will keep moving forward (Note some people describe this by saying that the person was thrown forward by the force of impact. Why is this wrong?). This means that the person will impact the steering wheel or windshield. The steering wheel and windshield can't collapse as nicely as the front of the car so Δt is much smaller and thus the average force is much higher.

   

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7. A lion of mass 120 kg leaps at a hunter with a horizontal velocity of 12m/s. The hunter has an automatic rifle firing bullets of mass 15 g with a muzzle speed of 630m/s and he attempts to stop the lion in midair. How many bullets would the hunter have to fire into the lion to stop its horizontal motion? Assume the bullets stick inside the lion.

   The total momentum of the system, the bullets and the lion, would be zero since there are no external forces to consider,
   

   P = m_lionv_lion - nm_bulletv_bullet = 0 ,
   

   where the lion is assumed to be moving in the positive direction. Rearranging the equation to find n,
   

   n = m_lionv_lion / m_bulletv_bullet = (120 kg 12 m/s) / (0.015kg 630 m/s) = 152.4 .
   

   It would take 153 bullets to stop the lion dead, so to speak.

   

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8. On a frictionless surface, a 6.0-kg rock approaches from the left at 3.5 m/s. It collides elastically with a 9.0-kg rock which is approaching from the right at 1.7 m/s. Find the final velocities of the rocks.

   We are dealing with a collision, so we know we must conserve momentum,
   

   m₁v_1f + m₂v_2f = m₁v_1i + m₂v_2i .           (1)

   We are told that the collision is elastic which means that kinetic energy is conserved. For a 1D collision, this is the same as
   

   v_2f - v_1f = -(v_2i - v_1i) .           (2)

   Substituting in the give data, our equations become

   6v_1f + 9v_2f = 6×3.5 + 9×(-1.7) = 5.7 ,
   v_2f - v_1f = -(-1.7 - 3.5) = 5.2 .

   Solving the two equations in two unknowns, we find v_1f = -2.74 m/s and v_2f = 2.46 m/s.
   

   

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9. If the collision in question #5 had been perfectly inelastic, what would have been the final velocity of the rocks? How much kinetic energy would have been lost in the collision?

   In any kind of collision, momentum is conserved so

   m₁v_1f + m₂v_2f = m₁v_1i + m₂v_2i .

   We are told that the collision is perfectly elastic which means that kinetic energy is not conserved and that the rocks stick together, v_1f = v_2f = v_f. Our equation becomes
   

   (m₁ + m₂)v_f = m₁v_1i + m₂v_2i .

   This allows us to find v_f immediately

   v_f = [m₁v_1i + m₂v_2i]/(m₁ + m₂) = [6 kg×3.5 m/s + 9 kg×(-1.7 m/s)]/ 15 kg = 0.38 m/s .

   The initial kinetic energy is
   

   K_i = ½m₁(v_1i)² + ½m₂(v_2i)² = ½(6)(3.5)² + ½(9)(-1.7)² = 49.755 J .

   The final energy is
   

   K_f = ½(m₁ + m₂)(v_f)² = ½(6+9)(0.38)² = 1.083 J .

   The difference in energy is
   

   ΔK = K_f - K_i = -48.7 J .

   

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10. If the collision in question #5 had a coefficient of restitution e = 0.600, what would have been the final velocity of the rocks? How much kinetic energy would have been lost in the collision?

    In any kind of collision, momentum is conserved so

    m₁v_1f + m₂v_2f = m₁v_1i + m₂v_2i .

    We are told that the coefficient of restitution

    v_2f - v_1f = -e(v_2i - v_1i) .

    Substituting in the give data, our equations become

    6v_1f + 9v_2f = 6×3.5 + 9×(-1.7) = 5.7 ,
    v_2f - v_1f = -(0.6)(-1.7 - 3.5) = 3.12 .

    Solving the two equations in two unknowns, we find v_1f = -1.492 m/s and v_2f = 1.628 m/s.
    

    The initial kinetic energy is
    

    K_i = ½m₁(v_1i)² + ½m₂(v_2i)² = ½(6)(3.5)² + ½(9)(-1.7)² = 49.755 J .

    The final energy is
    

    K_f = ½m₁(v_1f)² + ½m₂(v_2f)² = ½(6)(-1.492)² + ½(9)(-1.628)² = 18.605 J .

    The difference in energy is
    

    ΔK = K_f - K_i = -31.1 J .

    

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11. A 50.0-kg skater is traveling due east at 3.00 m/s. A 70.0-kg skater is moving due south at 7.00 m/s. They collide and hold on to one another after the collision. Determine the magnitude and direction of their velocity after the collision. Ignore the effects of friction.
    

    In any kind of free collision, momentum is conserved so

    (m₁ + m₂)v_f = m₁v_1i + m₂v_2i .           (1)

    Now momentum and velocity are vector quantities and the i and j components must be handled separately
    

    (m₁ + m₂)v_fx = m₁v_1ix + m₂v_2ix ,          (1a)
    (m₁ + m₂)v_fy = m₁v_1iy + m₂v_2iy .           (1b)

    So we can rearrange these equations to find the components of the final velocity
    

    v_fx = (m₁v_1ix + m₂v_2ix) / (m₁ + m₂) ,          (2a)
    v_fy = (m₁v_1iy + m₂v_2iy) / (m₁ + m₂) .           (2b)

    Using the given values, we find

    v_fx = [(50 kg)(3 m/s) + (70 kg)(0)] / (50 kg + 70kg) = 1.25 m/s ,           (2a)
    v_fy = [(50 kg)(0) + (70 kg)(-7 m/s) / (50 kg + 70 kg) = 4.083 m/s .         (2b)

    To find the magnitude and direction of the final velocity, we use Pythagoras' Theorem and trigonometry,
    

    v_f = [(v_fx)² + (v_fx)²]^½ = 4.27 m/s, and
    θ = tan^-1 (|v_fy/v_fx|) = 72.98°.

    The final velocity of the pair is 4.27 m/s at 73.0° south of east.

    

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12. Football player A tackles and holds onto player B in the diagram below.
    
    1. Assuming friction is negligible, what is the velocity of the players just after the collision?
    2. If the collision lasts for 0.30 seconds, what average force (magnitude and direction) does player A exert on player B?
    

    In any kind of free collision, momentum is conserved so

    (m_A + m_B)v_f = m_Av_Ai + m_Bv_Bi .           (1)

    Now momentum and velocity are vector quantities and the i and j components must be handled separately
    

    (m_A + m_B)v_fx = m_Av_Aix + m_Bv_Bix ,          (1a)
    (m_A + m_B)v_fy = m_Av_Aiy + m_Bv_Biy .           (1b)

    So we can rearrange these equations to find the components of the final velocity
    

    v_fx = (m_Av_Aix + m_Bv_Bix) / (m_A + m_B) ,          (2a)
    v_fy = (m_Av_Aiy + m_Bv_Biy) / (m_A + m_B) .           (2b)

    Using the given values, we find

    v_fx = [(75 kg)(2.75 m/s)cos(50°) + (70 kg)(2.30 m/s)] / (75 kg + 70kg) = 2.025 m/s ,           (2a)
    v_fy = [(75 kg)(2.75 m/s)sin(50°) + (70 kg)(0) / (75 kg + 70 kg) = 1.090 m/s .         (2b)

    To find the magnitude and direction of the final velocity, we use Pythagoras' Theorem and trigonometry,
    

    v_f = [(v_fx)² + (v_fx)²]^½ = 2.30 m/s, and
    θ = tan^-1 (|v_fy/v_fx|) = 28.3°.

    The final velocity of the pair is 2.30 m/s at 28.3° above horizontal.

    The force on player B has to do with the impulse acting on him and his change of momentum,

    F_average = I_B/Δt                     (3)

    Now force and impulse are vector quantities and the i and j components must be handled separately.

    F_{ave x} = (m_Bv_Bfx - m_Bv_Bix)/Δt ,                     (3a)

    F_{ave y} = (m_bv_Bfy - m_Bv_Biy)/Δt .                     (3b)

    Using the given values, we find

    F_{ave x} = (70 kg)[(2.025 m/s) - (2.30 m/s)] / (0.30 s) = -64.17 N ,           (4a)

    F_{ave y} = (70 kg)[(1.090 m/s) - (0)] / (0.30 s) = 254.33 N .         (4b)

    To find the magnitude and direction of the average force, we use Pythagoras' Theorem and trigonometry,
    

    F_ave = [(F_x)² + (F_y)²]^½ = 262 N, and
    θ = tan^-1 (|F_y/F_x|) = 75.8°.

    The average force on player B is 262 N at 94.2° above horizontal.

    

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13. A curling rock is traveling down the ice when it mysteriously explodes into three parts. After the explosion, one piece having 27.0% of the total mass moves at a speed of V_1f = 14.2 m/s at an angle of 42.0° to the positive y axis. A second with 52.0% of the total mass move at a speed of V_2f = 18.9 m/s at an angle of 17.8° to the positive x axis. The third piece moves with speed V_3f = 35.9 m/s at 39.0° to the negative y axis. What was the speed of the stone before the explosion?
    

    In any kind of collision, momentum is conserved so

    m₁V_1f + m₂V _2f + m₃V_3f = (m₁ + m₂ + m₃)V_i .           (1)

    Now momentum and velocity are vector quantities and the i and j components must be handled separately
    

    m₁V_1fx + m₂V_2fx + m₃V_3fx = (m₁ + m₂ + m₃)V_ix .           (1a)
    m₁V_1fy + m₂V_2fy + m₃V_3fy = (m₁ + m₂ + m₃)V_iy .           (1b)

    So we can rearrange these equations to find the components of the final velocity
    

    V_ix = (m₁V_1fx + m₂V_2fx + m₃V_3fx) / (m₁ + m₂ + m₃) ,           (2a)
    V_iy = (m₁V_1fy + m₂V_2fy + m₃V_3fy) / (m₁ + m₂ + m₃) .           (2b)

    Taking the masses of the piece to be m₁ = 0.27M, m₂ = 0.52M, and m₃ = (1-0.27-0.52)M = 0.21M where M is the total mass, and using the given values, we find
    

    V_ix = [0.27(-14.2sin(42°)) + 0.52(18.9cos(17.8°)) + 0.21(35.9sin(39°))] = 2.05 m/s ,
    V_iy = [0.27(14.2cos(42°)) + 0.52(18.9sin(17.8°)) + 0.21(-35.9cos(39°))] = 0.0 m/s .

    The stone had an initial speed of 2.05 m/s to the right.
    

    

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14. Two opposing hockey players are racing up the ice for the puck when they collide at point A as shown in the diagram below. The first hockey player has mass 90 kg and a speed of 2.7 m/s while the other has mass 82 kg and speed 3.1 m/s. The angle in the diagram is θ = 32° . After the collision, the players remain locked together (at least until the referee forces them apart). What is the magnitude and direction of the players' velocity just after they collide?
    

    Since the collision is totally inelastic and in two dimensions, we find that we are dealing with a vector addition problem, P_T = P₁ + P₂. First we calculate the magnitude of each player's momentum using p = mv,
    

    P₁ = m₁v₁ = (90 kg)(2.7 m/s) = 243 kg-m/s ,
    
    P₂ = m₂v₂ = (82 kg)(3.1 m/s) = 254.2 kg-m/s.
    

    Then we find P_T by the component method,

    i		j
    P1x	= 0	P1y	= 243
    P2x	= 254.2sin(32°) = 134.705	P2y	= 254.2cos(32°) = 215.574
    PTx	= 134.705	PTy	= 458.574

    Using the Pythagorean formula we find,
    

    P_T = [(P_Tx)² + (P_Ty)²]^½ = [(134.705)² + (458.574)²]^½ = 477.949 kg-m/s .
    

    Using trigonometry, we find the angle from
    

    θ = arctan(P_Ty/P_Tx) = arctan(458.574/134.705) = 73.63°.
    

    So the total momentum of the two players is P_T = (478,73.6°).

    Now P_T = (m₁ + m₂)v_f, so the final velocity must be in the same direction as the total momentum. The magnitude of the velocity is
    

    v_f = P_T/(m₁ + m₂) = 477.949 kg-m/s/ (90 kg + 82 kg) = 2.78 m/s .
    

    So the final velocity of the two players just after the collision is v_f = (2.78 m/s, 73.6°).

    

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15. The distance between the oxygen molecule and each of the hydrogen atoms in a water (H₂O) molecule is 0.0958 nm; the angle between the two oxygen-hydrogen bonds is 105°. Treating the atoms as particles, find the centre of mass.
    

    The problem expects you to recall that the mass of an oxygen atom is 16 times that of a hydrogen atom. The first step is to choose a coordinate system, such as the one in the diagram, and locate each particle. The chosen origin is the centre of the box.

    

    Atom	Mass (H)	xi	yi	mixi	miyi
    H	1	-0.0958sin15	0.0958cos15	-0.02479	0.09254
    O	16	0	0	0	0
    H	1	0.0958	0	0.0958	0
    Totals:	18			0.07101	0.09254

    The coordinates of the centre of mass are given by

    x_cm = (Σm_ix_i)/M_total = 0.07101/18 = 0.0039 nm, and
    y_cm = (m_iy_i)/M_total = 0.09254/18 = 0.0051 nm.

    The centre of mass is located at (0.0039 nm, 0.0051 nm). Answers will vary based on the choice of coordinate system.

    

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16. Where is the centre of mass of a uniform cubic box of side length L which has no lid?
    

    In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes is at their easy to find geometric centre if the object is uniform. Each pierce can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles. In this case, each of the five sides can be considered a separate particle.

    The next step is to choose a coordinate system, such as the one in the diagram below, and locate each particle. The origin is at the centre of the box.

    

    Side	Mass	xi	yi	zi	mixi	miyi	mizi
    bottom	M	0	0	-½L	0	0	-½ML
    front	M	0	-½L	0	0	-½ML	0
    back	M	0	½L	0	0	½ML	0
    left	M	-½L	0	0	-½ML	0	0
    right	M	½L	0	0	½ML	0	0
    Totals:	5M				0	0	-½ML

    The coordinates of the centre of mass are given by
    

    x_cm = (Σm_ix_i)/M_total = 0,
    y_cm = (Σm_iy_i)/M_total = 0, and
    z_cm = (Σm_iz_i)/M_total = -½ML / 5M = -L/10.

    The centre of mass is located at (0, 0, -L/10). Answers will vary based on the choice of coordinate system.

    It is also permissible to use symmetry arguments. For example, the figure in the diagram is only unbalanced in the z direction, thus we know that x_cm = y_cm = 0. We only needed the z columns in the above table.
    

    

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17. Two uniform squares of sheet metal of dimension L × L are joined at a right angle along one edge. One of the squares has twice the mass of the other. Find the centre of mass.
    

    In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes is at their easy to find geometric centre if the object is uniform. Each pierce can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles. In this case, we have one particle of mass M located in the centre of the lighter side, and a mass of 2M in the centre of the heavier side.

    The next step is to choose a coordinate system, such as the one in the diagram below, and locate each particle. The origin is in the centre of the join of the two plates.

    

    Using the symmetry of the problem, we see that the CM must be located in the xz plane, we know that y_cm = 0.

    Side	Mass	xi	zi	mixi	mizi
    side	M	0	½L	0	½ML
    bottom	2M	½L	0	ML	0
    Totals:	3M			ML	½ML

    The components of the centre of mass are given by

    x_cm = (Σm_ix_i)/M_total = ML / 3M = L/3,
    z_cm = (Σm_iz_i)/M_total = ½ML / 3M = L/6.

    The centre of mass is located at (L/3, 0, L/6). Answers will vary based on the choice of coordinate system.

    

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18. A cube of iron has dimension L × L × L. A hole of radius ¼L has been drilled all the way through the cube, so that one side of the hole is tangent to one face along its entire length. Where is the centre of mass of the drilled cube?
    

    In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes is at their easy to find geometric centre if the object is uniform. Each pierce can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles. In this case, we have a solid cube and a cylindrical hole. We treat holes as objects of negative mass.
    

    To proceed we need to know the mass of the cylindrical hole. Since the object was uniform, its mass is proportional to its volume. The solid cube had a mass M and a volume L³. The cylinder has a volume V_cyl = r²L = L³/16. Thus the mass of the cylindrical hole is

    m_cyl = m_cube[V_cyl / V_cube] = M[(πL³/16) / L³] = πM/16.

    The next step is to choose a coordinate system, such as the one in the diagram below, and locate each particle.

    

    Using the symmetry of the problem, we see that the CM must be located in the x axis, we know that y_cm = z_cm = 0.

    Side	Mass	xi	mixi
    solid cube	M	0	0
    hole	-πM/16	-¼L	πML/64
    Totals:	M(1-π/16)		πML/64

    The location of the x component of the centre of mass is given by

    x_cm = (Σm_ix_i)/M_total = -πML/64 / -M(1-π/16) = πL / 64(1-π/16).

    The centre of mass is located at (πL / 64(1-π/16), 0, 0). Answers will vary based on the choice of coordinate system.

    

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19. An 80-kg logger is standing on one end of a 10-m long, 300-kg, tree trunk in the middle of the Fraser River. The logger walks upriver along the trunk to the other end of the log. As a result the log moves some distance L down river. What is the displacement L?
    

    The logger and the log are a system; the system has a certain centre of mass. The motion of the logger is an internal force; internal forces cannot change the centre of mass of the system.

    

    Examining the diagram, the log has moved down the river a distance equal to twice the distance between the centre of the log and the CM of the logger-log system. We need to find the CM of the logger-log system. Taking the origin as the end of the log,

    x_cm = (Σm_ix_i)/M_total = [80×0 + 300×5] / 380 = 3.047 m.

    Since the centre of the log is at 5 m, distance between the CM and the centre of the log is 1.05 m. The log moved twice this distance or 2.11 m.

    

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20. A shell is fired at 25 m/s at 25 above the horizontal. At the top of its parabolic flight, it breaks into two pieces. One piece, having two-thirds of the total mass of the shell lands 60 m from where the shell was fired. Where did the other piece land?

    The pieces of the shell are a system; the system has a certain centre of mass. The explosion is an internal force; internal forces cannot change the centre of mass of the system. The CM of the pieces will land where the CM of an unexploded shell will land.

    

    The first step is to find x_cm, the landing position of the shell. That involves solving the projectile motion problem.

    i	j
    x = xcm = ?	y = 0
    ax = 0	ay = -g = -9.81 m/s
    v0x = 25cos25 = 22.658 m/s	v0y = 25sin25 = 10.565 m/s
    t = ?	t = ?

    The j information allows us to find the time in air using y = v_0yt - ½gt². Since y = 0, this reduces to
    

    t = 2v_oy/g = 2×10.565 / 9.81 = 2.154 s.

    We then find the landing position using x = v_0xt + ½a_xt². Since a_x = 0,

    x_cm = v_oxt = 22.658 × 2.154 = 48.806 m.

    The centre of mass is determined by the formula

    x_cm = (Σm_ix_i)/M_total = m₁x₁/M_total + m₂x₂/M_total.

    Since we now know x_cm and x₂, we can rearrange to find m₁,

    x₁ = [M_totalx_cm - m₂x₂]/m₁ = [1×48.806 - (2/3)60] / (1/3) = 26.4 m.

    So the smaller piece lands 26.4 m from where the shell was fired.

    

Questions? mike.coombes@kwantlen.ca