collision problem

[anum.ahmed8882]

a 10kg block starts from rest at a height of 20m and slides down a frictionless semicircular track, the block collides with a stationary object of 50kgat the bottom of the track. Determne the height reached by each object if the collision was inelastic and the falling mass rebounded back with a speed of 1m/s

[calena7178147]

Btw, what you will see below is far more calculating than you will see on a real MCAT test. Nonetheless, just for fun . . . !

Let us first calculate the velocity of mass M after collision. It is an inelastic collision, so kinetic energy will not be conserved. However, Momentum will be.

So, velocity of mass m, before collision, after sliding down through a height of 20 m, is -
m g h = 0.5 m v^2 (no energy lost as the track is frictionless).

v^2 = 2 x g x h
With, g = 9.80 m s^-2, and h = 20 m
v^2 = 2 x 9.80 x 20 = 392
v = 19.8 m/ s


Applying Conservation of Momentum
Total Initial Momentum = Total Final Momentum
m v + M u = m v1 + M v2 .............................Eq. 1
where,
m = 10 kg mass of the sliding block
v = 19.8 m/ s velocity of the sliding block before colliding with the stationary mass M
M = 50 kg mass of the stationary object at the bottom of the track
u = 0 m/ s velocity of the stationary mass M before collision
v1 = - 1 m/ s velocity of the mass m after colliding, remember it rebounds so the –
v2 = ? velocity of the mass M after colliding

Substituting in Equation 1
10 x 19.8 + 50 x 0 = 10 x (- 1) + 50 x v2
198 = - 10 + 50 x v2
v2 = 208/ 50
v2 = 4.16 m/ s
Velocity of mass m after collision = v1 = - 1m/ s
Height reached by mass m
m g h1 = 0.5 x m x v1^2
h1 = v1^2/ (2 x g)
h1 = (-1)^2/ (2 x 9.80)
h1 = 0.051 m = 5.1 cm
Velocity of mass M after collision = v2 = 4.16 m/ s
Height reached by mass M
M g h2 = 0.5 x M x v2^2
h2 = v2^2/ (2 x g)
h2 = 4.16^2/ (2 x 9.80)
h2 = 0.88 m = 88 cm