Physics 101: Calculating Velocities of Thrower and Catcher

[jaredgeorge19]

A 66.0 kg person throws a 0.0400 kg snowball forward with a ground speed of 28.0 m/s. A second person, with a mass of 57.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.

a.thrower

b.catcher

 

[tiny-tim]

Hi jaredgeorge19! Welcome to PF!

Show us how what you've tried, and where you're stuck, and then we'll know how to help!

 

[kerimek]

I would approach this problem by first considering the conservation of momentum and the law of conservation of energy. The initial momentum of the system, consisting of the thrower and the snowball, is equal to the final momentum of the system, consisting of the catcher and the snowball. The initial kinetic energy of the system is also equal to the final kinetic energy of the system.

Using the equation for conservation of momentum, I can calculate the velocity of the catcher after the snowball is exchanged. This can be done by setting the initial momentum equal to the final momentum:

m1v1 + m2v2 = m1v1' + m2v2'

Where m1 and v1 are the mass and initial velocity of the thrower, m2 and v2 are the mass and initial velocity of the snowball, and v1' and v2' are the velocities of the thrower and catcher after the exchange.

Plugging in the given values, we get:

(66.0 kg)(2.20 m/s) + (0.0400 kg)(28.0 m/s) = (66.0 kg)(v1') + (57.0 kg)(v2')

Solving for v2', we get:

v2' = [(66.0 kg)(2.20 m/s) + (0.0400 kg)(28.0 m/s) - (66.0 kg)(v1')] / (57.0 kg)

v2' = 2.53 m/s

Therefore, the velocity of the catcher after the snowball is exchanged is 2.53 m/s.

To calculate the velocity of the thrower after the exchange, we can use the equation for conservation of energy:

KE1 + KE2 = KE1' + KE2'

Where KE1 and KE2 are the initial kinetic energies of the thrower and the snowball, and KE1' and KE2' are the final kinetic energies of the thrower and the catcher.

Since the thrower and the snowball are initially moving together, their total kinetic energy is:

KE1 + KE2 = (1/2)(66.0 kg + 0.0400 kg)(2.20 m/s)^2 = 167.0 J

After the exchange, the thrower is moving with a velocity of v1', so their kinetic energy is:

KE1' = (1/2)(