Question

I am getting 2.8 m/s, but my friend got 3.2 m/s who is right, or what am I doing wrong?

A package of mass M is releasd from rest as a warehouse loading dock and slides down a 3.6 m high frictionless chute. At the bottom of the chute, it hits the second package of mass 2M that was at rest. What is the common speed (in m/s) of the two packages after a perfectly inelastic collision?

Thanks!

Answer 1

A package of mass M is releasd from rest as a warehouse loading dock and slides down a (h) =3.6 m high frictionless chute. At the bottom of the chute, it hits the second package of mass 2M that was at rest. What is the common speed (in m/s) of the two packages after a perfectly inelastic collision.

Then the velocity of the package with mass M just before it hits the other package is given by

v =Sqrt(2gh) =Sqrt(2*9.81*3.6) =8.404m/s

After this at the bottom of the chute, it hits the second package of mass 2M that was at rest, then the common speed (in m/s) of the two packages after a perfectly inelastic collision is given by

M(8.404m/s)+2M*(0) =(M+2M)Vf

M(8.404m/s) =(3M)Vf

Then common speed is Vf =8.404m/s/3 =2.8014m/s

Your answer is correct.