Question

# In the figure, a 4.5 kg box of running shoes slides on a horizontal frictionless table...

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?

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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