Answer to Question #106027 in Mechanics | Relativity for Valentine Agun

Question

A 65.0 - kg person throws a 0.045 0 - kg snowball forward with a ground speed of 30.0 m/s. A second person, with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.50 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

Answer 1

We first consider the first person, which has a mass $m_1=65.0kg,$ and moves initially with a speed $v_{1f}=2.50m/s$ holding a snowball of mass $m_b=0.045kg.$ This person, together with the ball, has an initial linear momentum

$p_{1i}=(m_1+m_b)v_{1i}.$

This person then throws the ball with speed $v_b=30.0m/s.$ The final linear momentum of the person and the ball, after the event of throwing, is

$p_{1f}=m_1v_{1f}+m_bv_b.$

Since there is no friction and the event of throwing the snowball involves forces internal to the system, we assume linear momentum is conserved and write $p_{1i}=p_{1f}$ that is,

$(m_1+m_b)v_{1i}=m_1v_{1f}+m_bv_b$

which can be immediately solved for the final velocity of the person:

$v_{1f}=\frac{(m_1+m_b)v_{1i}−m_bv_b}{m_1}$

=$\frac{(65.0kg+0.045kg)(2.50m/s)−(0.045kg)(30.0m/s)}{65.0kg}$

$≈2.48m/s.$

Now, a second person, of mass $m_2=60.0kg,$ initially at rest, catches the snowball. Considering now the system snowball + second person, and neglecting the air resistance and gravity effects, so that the snowball arrives at the second person with the same speed it was thrown, we can write the initial momentum as

$p_{2i}=m_bv_b$

and the final momentum as

$p_{2f}=(m_b+m_2)v_{2f}$

.

The act of catching the ball involves internal forces, so the momentum is conserved and we have $p _{2i}=p_{2f}$ or

$m_bv_b=(m_b+m_2)v_{2f}$

The velocity of the second person is

$v_{2f}=\frac{m_bv_b}{m_b+m_2}$

$=\frac{(0.045kg)(30.0m/s)}{0.045kg+60.0kg}$

$≈0.022m/s.$