Starting at rest, two students stand on 11.5-kg sleds, which point away from eac

Question

Starting at rest, two students stand on 11.5-kg sleds, which point away from each other on ice, and they pass a 5.50-kg medicine ball back and forth. The student on the left has a mass of 55.7 kg and can throw the ball with a relative speed of 10.7 m/s. The student on the right has a mass of 42.7 kg and can throw the ball with a relative speed of 13.5 m/s. (Assume there is no friction between the ice and the sleds and no air resistance.)

a) If the student on the left throws the ball horizontally to the student on the right, how fast is the student on the left moving right after the throw?

b) How fast is the student on the right moving right after catching the ball?

c) If the student on the right passes the ball back, how fast will the student on the left be moving after catching the pass from the student on the right?

d) How fast is the student on the right moving after the pass?

Explanation / Answer

Starting at rest, two students stand on 11.9-kg sleds, which point away from each other on ice, and they pass a 4.50-kg medicine ball back and forth. The student on the left has a mass of 57.9 kg and can throw the ball with a relative speed of 10.9 m/s. The student on the right has a mass of 42.5 kg and can throw the ball with a relative speed of 12.3 m/s. (Assume there is no friction between the ice and the sleds and no air resistance.)

If the students were standing on solid ground, they could throw the ball at 10.9m/s and 12.3 m/s respectively. They aren't on the ground though; they are on these sleds. If you have ever fallen off a skateboard and had the skateboard fly forward as you went backwards, you know what this is like.
(a) there is zero momentum before the left student throws the ball and conservation of momentum tells us that there is zero afterwards. So the masses of the left student and the left sled times their speed equals the mass of the ball times its speed. The difference in velocity is 10.9 m/s. We will use algebra and say left students speed is x and the ball's speed is 10.9 - x (we will use algebra again in part c)
(57.9+11.9) x = 4.5 (10.9 -x)
69.8x + 4.5x = 49.05
x = 49.05/74.3 = 0.660161507
10.9 - x = 10.23983849
The student on the left is moving at 0.66 m/s to the left.
Check this answer 0.66+10.24 ?=? 10.9 Yes! the left student sees the ball leave his hands at 10.9 m/s.

(b) The ball is traveling at 10.24m/s to the right when the right student catches it. Momentum of the ball and the right student and sled is equal to the momentum of the ball
4.5(10.24) = (4.5+11.9+42.5)V
V = 4.5(10.24)/(4.5+11.9+42.5) = 0.782330615
So the student on the right is moving to the right at 0.78 m/s

(c) Student on the right passes the ball back. He is already moving to the right, and when he throws the ball, he will see it leave his hands at 12.3 m/s, but again it will push him with the same force that he pushes it, so that will just be the relative speed. There are two ways we could do this. (1) We could temporarily define a reference frame that is moving at 0.78m/s to the right and call this 0m/s. Afterwards we would add 0.78 m/s to our answer. (2) Or we could use the ground as our reference frame and solve it directly.

Method (1) 4.5(12.3-x) = (11.9+42.5)x
55.35 - 4.5x = 54.4x
58.9 x = 55.35
x = 55.3/58.9 = 0.939728353 -this is in our moving reference frame. To translate this to the ground reference frame we need to add 0.78 m/s to the right to our results.
So the right student is moving right at (0.939728353 + 0.78) m/s - this is the answer to part (d)
and the ball is moving left at (12.3 - 0.939728353 - 0.78) = 10.56 m/s
Why is it OK to change the reference frame? Momentum has velocity to the first power so we can just adjust the zero without changing the relative momentum (can't reset the zero when finding kinetic energy, since it has v^2)
The student on the left catches the ball.
He was traveling 0.66 m/s to the left.
Let's solve this one directly
Momentum is conserved.
Left student and his sled have momentum before the collision plus ball's momentum equals momentum of left student, sled, and ball after the collision.
(11.9+42.5)0.66 + 4.5*10.56 = (4.5+11.9+42.5)V
V = [(11.9+57.9)0.66 + 4.5*10.56]/(4.5+11.9+57.9) = 1.260834561
so the student on the left (and the ball and the sled) will be traveling to the left at
v = 1.26 m/s
(d) the student on the right will be traveling at (0.939728353 + 0.78) m/s = 1.72 m/s to the right as we found above when we calculated the speed of the ball to the left.

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