Question

A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to a waiting truck. It happens a box of m2 = 40.0 kg was left at the bottom of the chute. What is the impulse acting on m2 during the collision between the two boxes?

Answer 1

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass $m_1$ just before the collision

$v_1 = \sqrt{2gh}$

$v_1 = \sqrt{2(9.8)(3)}$

$v_1 = 7.67m/s$

Therefore the momentum just before collision would be

$p_2 = m_1v_1+40(0)\\p_1 = 20*7.67+40(0)\\p_1 = 153.36kg \cdot m/s$

Momentum after the collision

$p_1 = 20*u_1+40u_1\\p_1 = 60u_1$

Since the momentum is conserved we have that

$153.36= 60u_1$

$u_1 = \frac{153.36}{60}$

$u_1 = 2.56m/s$

The velocity of mass $m_2$ after the collision is given by

$v_2 = \frac{2m_1}{m_1+m_2} u_1$

$v_2 = \frac{2(20)}{20+40}(2.56)$

$v_2 = 1.71m/s$

Therefore the change in momentum of mass 2 is

$p_2 = m_2v_2$

$p_2 = 40*1.71$

$p_2 = 68.4kg\cdot m/s$

Therefore the impulse acting on m2 during the collision between the two boxes is $p = 68.4kg\cdot m/s$