m₁=2.4 kg, m₂=1.4 kg, v₁=4.6 m/s, v₂=0
m₁v₁= (m₁+m₂) u
u= m₁v₁/(m₁+m₂) =2.4•4.6/(2.4+1.4) =2.9 m/s
At the edge of the table, two boxes have
kinetic (KE) and potential energy (PE), and near the ground - KE₁
KE + PE=KE₁
(m₁+m₂)u²/2 + (m₁+m₂)gh =(m₁+m₂)v²/2
u²/2 +gh = v²/2
v=sqrt{ u²+2gh} =sqrt{2.9² +2•9.8•0.75}=3.65 m/s
In the figure, a 2.4 kg box of running shoes slides on a horizontal frictionless table and collides with a 1.4 kg box of ballet slippers initially at rest on the edge of the table, at height h = 0.75 m. The speed of the 2.4 kg box is 4.6 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?
1 answer
1 answer