Question

A 67.5-kg person throws a 0.0410-kg snowball forward with a ground speed of 34.0 m/s. A second person, with a mass of 57.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.45 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged?

Answer 1

Answer:

Final velocity of the first person is 3.43m/s and that of the second person is 0.0242m/s

Explanation:

Let the momentum of the first person, the ball second person be Ma, Mb and Mc.

From the principle of the conservation of momentum, sum of the momentum before collision is equal to the sum of the momentum after collision.

Ma1 + Mb1 = Ma2 + Mb2.

The ball and the first person are both moving together with a common velocity 3.45m/s.

Let the velocity of the first person be v1

Therefore

67.5×3.45+ 0.041×3.45= 67.5v1 + 0.041×34

233.02 = 1.39+ 67.5v1

67.5v1 = 233.02 - 1.39 = 231.61

v1 = 231.61 / 67.5

v1 = 3.43m/s

The second person and the ball move together with a common velocity after catching the ball.

For the second person and the ball let their final common velocity be v

Mb2 + Mc2 = Mb3 + Mc3

0.041 × 34 + 57.5 ×0 = (57.5 + 0.041)×v

57.541v = 1.39

v = 1.39 /57.541

v = 0.0242m/s