Question

In the figure below, a 3.9 kg box of running shoes slides on a horizontal frictionless table and collides with a 2.4 kg box of ballet slippers initially at rest on the edge of the table, at height h = 0.40 m. The speed of the 3.9 kg box is 5.0 m/s just before the collision. If the two boxes stick together because of packing tape on their sides, what is their kinetic energy just before they strike the floor?

Answer 1

Using Law of Conservation of momentum

Momentum before collision = momentum after collision

$3.9 kg \times 5.0 m/s+2.4 kg\times 0 m/s = (3.9 kg + 2.4 kg )\times v_{x}$

$v_{x} = \frac{3.9 kg \times 5.0 m/s}{(3.9 kg + 2.4 kg )}= 3.095 m/s$

This will remain horizontal velocity at the time of collision. Now vertical velocity gain during fall

$v_{y} = \sqrt{2gh} = \sqrt{2\times 9.8 m/s^{2}\times 0.40 m}= 2.8 m/s$

Velocity of the system just before collision

$v = \sqrt{v_{x}^{2}+v_{y}^{2} }= \sqrt{(3.095 m/s)^{2}+(2.8 m/s)^{2}} =4.174 m/s$

Kinetic energy before striking floor

$KE = \frac{1}{2}(m_{1}+m_{2})v ^{2}=\frac{1}{2}\times (2.4kg +3.9 kg)\times (4.174 m/s)^{2}= 54.87 J$