Question

In the figure, a 4.5 kg box of running shoes slides on a horizontal frictionless table and collides with a 2.8 kg box of ballet slippers initially at rest on the edge of the table, at height h = 0.60 m. The speed of the 4.5 kg box is 4.8 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 Energy

Energy at the top = Energy at the bottom

0.5*(m1+m2)*V^2 + (m1+m2)*g*h = 0.5*(m1+m2)V2^2

m1+m2 cancels on both sides

0.5*V^2 + (g*h) =0.5*v2^2

using law of conservation of linear momentum

Momentum of the system before collision = momentum of the system
after the collision

m1*u = (m1+m2)*V

4.5*4.8 = (4.5+2.8)*v

v = 2.96 m/s

then

0.5*V^2 + (g*h) =0.5*v2^2

(0.5*2.96^2)+(9.8*0.6) = 0.5*v2^2

v2 = 4.53 m/s

Kinetc energy is = 0.5*(m1+m2)*v2^2 = 0.5*(4.5+2.8)*4.53^2 =
**74.9 J**

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