Question

Two physics students are standing on skateboards and facing each other. Their masses, including the skateboards, are 95 kg and 58 kg. Both students are initially at rest. Then they are pushing off each other, and the larger student moves with a speed of 3.1 m/s.
1) What is the total momentum of the system consisting of the two students after they push off? (ignore any friction)



2) What is the speed of the ligther student after they push off?

Answer 1

Answer:

1) $\sum \vec p=0\ kg.m.s^{-1}$

2) $v_2=55.078\ m.s^{-1}$

Explanation:

Given:

• mass of heavier student, $m_1=95\ kg$

• mass of lighter student, $m_2=58\ kg$

• velocity of the heavier student after the mutual push, $v_1=3.1\ m.s^{-1}$

1)

Since they push each other and we are neglecting any friction that means we can consider this as a perfectly elastic collision. So the momentum of the two bodies will be equal and in opposite direction.

Since momentum is a vector quantity so the total momentum of the system as a whole will be zero.

2)

From the condition of elastic collision:

$p_1=p_2$

$m_1.v_1=m_2.v_2$

$95\times 3.1=58\times v_2$

$v_2=55.078\ m.s^{-1}$