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Homework Assignment 4
Due: Monday, June 13, 2016
Problem 1: A lion of mass 150 kg and a grizzly bear of mass 300 kg are charging directly towards each other at 75 km/h and 40 km/h respectively. Find the magnitude of momentum of
(a) the lion.
(b) the bear.
(c) the lion-bear system.
Solution
Let the direction of the lion’s motion be positive, and that of the bear be negative. The speeds of the lion and bear in m/s are respectively 20.83 m/s and 11.11 m/s.
(a) plion = mlionvlion = 3124.5 kg m/s or 11250 kg km/hr
(b) pbear = mbearvbear = −3333.3 kg m/s or −12000 kg km/hr. The minus sign will be dropped in the magnitude.
(c) psystem = plion + pbear = −208.8 kg m/s or −750 kg km/hr. Again, the minus sign will be dropped in the magnitude.
Problem 2: A tennis ball of mass 60 g hits a wall horizotally at a speed of 30 m/s and rebounds with the same speed. Just before and just after collision, the ball is traveling horizontally.
(a) What is the magnitude of momentum of the ball just before collision? What is the magnitude of momentum of the ball just after collision?
(b) What is the magnitude of the change in momentum of the ball?
(c) If the collision between the wall and the ball lasts for 0.1 s, what is the magnitude of average force during collision?
Solution
Let the direction in which the ball is traveling before collision be positive, and the direction after collision be negative. This choice won’t affect any of the answers because we’re only interested in the magnitudes.
(a) pi =. 060 × 30 kg m/s = 1 .8 kg m/s. pi =. 060 × (−30) kg m/s = − 1 .8 kg m/s.
(b) ∆p = pf − pi = − 3 .6 kg m/s.
(c) Using impulse-momentum theorem ∆p = Fav∆t, we get Fav = −36 N. The minus signs will be dropped in the magnitudes for all three parts.
Problem 3: A person of mass 70 kg is standing on an icy (frictionless) horizontal surface with a metal ball of mass 5 kg. The person throws the ball at a speed of 2 m/s, at an angle of 60◦^ above the horizontal. Both speed and angle are as measured with respect to the ground.
(a) What is the recoil speed with which the person slides after throwing the ball?
(b) What is velocity of the ball as seen by the person? Give the magnitude and direction (angle above horizontal).
Solution
(a) In this problem, only the x-component of momentum is conserved in the collision (the throw). The person’s velocity in the y-direction is 0 throughout. Because there are relative velocities invloved, we’ll be carefull with indices in our solution (B for ball, P for person, and G for the ground). We’ll do away with the index for initial and final; all velocities in the solution are after the throw. We’re given that the velocity of the ball with respect to the ground is 2 m/s at an angle of 60◦^ above the horizontal. Let’s pick the x-axis along the horizontal component of the ball’s velocity. Then,
vxBG = 2 cos 60◦^ m/s = 1 m/s vyBG = 2 sin 60◦^ m/s = 1.73 m/s
vP Gy = 0 vP Gx =?
From conservation of momentum in the x-direction we have
mB vBGx + mP vxP G = 0 (1)
Solving this we get vP Gx = − 0 .07 m/s, -0.071 m/s if we include an extra sig-fig. This is the recoil velocity of the person; the magnitude will hae the minus sign dropped.
(b) Now we use ~vBP = ~vBG − ~vP G (2) to find
vxBP = 1 − (− 0 .07) = 1.07 m/s vyBP = 1. 73 − 0 = 1.73 m/s
This corresponds to a magnitude of 2.04 m/s and an angle of 58. 26 ◦. This is very close to the velocity with respect to the ground, because the mass of the person is much larger than the ball’s, so the recoil is relly slow.
Problem 6: A two object system is known to have its center of mass moving at 3 m/s in the +x-direction. The two components have masses M and M/2. The heavier component is moving at 4 m/s at an angle of 60 ◦^ above the +x-axis. Find the velocity (magnitude and direction) of the lighter compenent.
Solution
The given velocities are to be plugged into the equation for the velocity of center of mass, to find the velocity of the lighter component. Since velocity is a vector, this needs to be done separately to the x and y components of velocity. The same applies to the law of conservation of momentum as well, if the momentum is conserved in more than one dimension. Let the heavy object be object #1, and the lighter object be object #2. The equations for the x and y components of the velocity of the center of mass give
3 m/s =
M × 4 cos 60◦^ m/s + M 2 v 2 ,x M + M 2
0 =
M × 4 sin 60◦^ m/s + M 2 v 2 ,y M + M 2
Solving these, we get v 2 ,x = 5 m/s and v 2 ,y = − 6 .93 m/s. This corresponds to a magnitude of 8.54 m/s and a direction of − 54. 18 ◦^ or 54. 18 ◦^ below the x-axis. Note, that in solving for v 2 ,x and v 2 ,y , M cancel out.
Problem 7: On a pool table, the cue ball, traveling at a speed of v, hits an object ball at rest head on (one dimensional collision). The cue ball and the object ball have the same mass. After collision, the cue ball keeps moving forward at a third of its original speed. What is the speed of the object ball after collision?
Solution
Conservation of momentum in one dimension:
vc,i = v vo,i = 0
vc,f =
v
vo,f =?
mcvc,i + movo,i = mcvc,f + movo,f
Since mc = mo, we can cancel out the masses. Solving this equation, we get vo,f = 23 v.
Problem 8: A car is traveling down a road at 50 m/s when Captain America latches onto it from the side of the road. If Captain America’s mass is 100 kg, and the speed of the car immediately after he gets on it is 45 m/s, what is the mass of the car (without including Captain America)? Assume that Cap was at rest on the ground before latching on.
Solution
One dimensional momentum conservation again. This time we’re solving for the mass of one of the objects.
mcar × 50 + mcap × 0 = (mcar + mcap) × 45 ⇒ mcar × (50 − 45) = mcap × 45 = 4500 kg
⇒ mcar =
kg = 900 kg
Problem 9: On a pool table, the cue ball traveling at speed of 2 m/s in the +x-direction hits an object ball at rest. The cue ball and the object ball have the same mass. After collision, the x-component of the cue ball’s velocity is 0.5 m/s, and the y-component of the object ball’s velocity is
√ 3 2 m/s. Find the velocities of the cue ball and object ball (magnitude and direction). What is the angle between the velocity vectors of the cue ball and object ball after collision?
Solution
Two dimension conservation momentum, this time. Conservation of momentum in the x-direction gives vxo,f = 1.5 m/s; and in the y- direction it gives
vyc,f = −
√ 3 2 m/s. Converting to magnitudes and angles, we find that the cue ball travels at 1 m/s at an angle of − 60 ◦; and the object ball travels at
3 m/s at an angle of + 30 ◦. The angle between their final trajectories is 90 ◦.
Problem 10: Two objects A and B collide. Just before collision the two objects are moving only in the x-direction (either +x or -x direction). Just after collision, object A keeps moving in the x-direction (either +x or -x direction). Can object B have a non-zero compenent of velocity along the y-direction just after collision? If yes, give an example. If no, explain why not.
Solution
Before collision, the total momentum in the y-direction is 0. So by conservation of momentum, after collision as well the total momentum in the y-direction should be 0. This means that if B has a non-zero component of velocity in the y-direction, then A’s velocity will need to have some component in the y-direction as well, so that the momenta can can out (in the y-direction); a contradiction since A is given to be moving only in the x-direction even after collision. So, B cannot have a non-zero y-component to its velocity. To explain with an equation, mAvy A,i + mBv By,i = mAvy A,f + mBvy B,f
If vy A,i, vy B,i, and vy A,f , then so is vy B,f (irrespective of what the masses of the objects are, and what the angle of contact is between the objects colliding). This was the expected answer. And all this is true unless, as some of you have noted, in the problem the momentum of the ‘A+B system’ isn’t conserved in the y-direction.
