Question

block 1 of mass m1 slides from rest along a frictionless ramp from an unknown height h and then collides with stationary block 2, which has mass m2 = 3m1 . The collision is an elastic one. After the collision, block 2 slides into a friction-filled region where the coefficient of kinetic friction is 0.5 and comes to a stop through a distance d = 10 m in that region. What is the height h?

Answer 1

m1 , h
m2 = 3 * m1
Elastic Collison !!

uk = 0.5
d = 10 m
acceleration in the friction-filled region, = uk*g = 0.5 * 9.8 = 4.9 m/s^2

Let the initial velocity be v.
vf^2 = v^2 - 2*a*d
0 = v^2 - 2*4.9*10
v = 9.9 m/s

Now using Energy conservation,
Initial Potential Energy of Block 1 = Final kinetic Energy of block 1
m1*g*h = 1/2*m1*vi^2
9.8*h = 1/2 * vi^2
vi^2 = 19.6*h -----1

Now, Uisng Momentum conservation,
Initial Momentum = Final Momentum
m1*v = m1*v1f + m2*v
m1*sqrt(19.6*h) = m1 * v1f + 3*m1*9.9
sqrt(19.6*h) = v1f + 29.7 ------1

As the collison is elastic, energy is conseved
Initial kinetic Energy = Final kinetic Energy
1/2 * m1 * vi^2 = 1/2 * m1 * v1f^2 + 1/2 * m2 * v^2
m1*19.6*h = m1*v1f^2 + 3*m1*9.9^2
19.6*h = v1f^2 + 294.03 --------2

Solving the two eq,
h = 20 m