Question

Two blocks are free to slide along a frictionless, wooden track. The track consists of a curve of height 4.91 m that falls to a straightaway. A smaller block weighing 4.91 kg is allowed to fall from the top the curve. It has a spring pointed forward embedded at the front. After reaching the straightaway, it collides with a larger block weighing 12 kg that is hard and flat at its rear. The two blocks rebound elastically. What is the final speed of the larger block?

Answer 1

we use spring, which here isn't really relevant -- they could be magnets, springs, or the blocks could actually collide (elastically).

Kinetic Energy = (1/2) mv² .
Change in potential energy = mgh .
Equating both.we get

(1/2)mv² = mgh
v² /2 = gh
v² = 2gh
v = √2gh

Also decrease in potential energy is equal to increase in kinetic energy

So, block(1) m=4.91kg comes into the "collision" with velocity
v = √(2gh)

For an elastic, head-on collision, we know that
the relative velocity of approach = relative velocity of separation, or
v = V - U
where V is the post-collision velocity of block(2) M=12kg
and U is the post-collision velocity of block(1) m
So
U = V - v

Now conserve momentum:
m(v) + M(0) = m(U)+ M(V)

m(V - v) + MV = (m+M)V - mv
2mv = (m+M)V
V = 2mv / (m+mM)

Plug in m, M and v = √(2gh)

where h = 4.91m
V = {2 × 4.91kg × √(2×(9.8m/s²) ×4.91m)} /{(4.91+ 12)kg}

V = 19.61m/s

So the block(2) which move with 19.61m/s