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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?

User M Falanga
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1 Answer

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´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 = √(2gh)


v_1 = √(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 = (153.36)/(60)


u_1 = 2.56m/s

The velocity of mass
m_2 after the collision is given by


v_2 = (2m_1)/(m_1+m_2) u_1


v_2 = (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

User Vlad Yarovyi
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