Question

A block of mass m1 = 1 kg is initially at rest at the top of an h1 = 1 meter high ramp, see Fig. 2 below. It slides down the frictionless ramp and collides elastically with a block of unknown mass m2, which is initially at rest. After colliding with m2, mass m1 recoils and achieves a maximum height of only h2 = 0.33 m going back up the frictionless ramp. (HINT: Solving each part in sequence will guide you to a solution for the value of m2, without doing a lot of algebra.)

a) Considering the energy of mass m1 just before and after the elastic collision, how much energy is lost by mass m1?
b) What is the speed of mass m1 just before the elastic collision?
c) What is the speed of mass m1 just after the elastic collision?

d) Considering conservation of momentum, what is the momentum of mass m2 just after the elastic collision?
e) What is the kinetic energy of mass m2 just after the elastic collision?
f) What is the velocity of mass m2 just after the elastic collision?

g) What is the value of mass m2 (in kg)?

Answer 1

a) Energy lost by m1 = m1*g*h1 - m1*g*h2

= m1*g*(h1 - h2)

= 1*9.8*(1 - 0.33)

= 6.566 J   <<<<<<<<<-----------------Answer

b) the speed of mass m1 just before the elastic collision, u1 = sqrt(2*g*h1)

= sqrt(2*9.8*1)

= 4.43 m/s <<<<<<<<<-----------------Answer

c) the speed of mass m1 just after the elastic collision, v1 = sqrt(2*g*h2)

= sqrt(2*9.8*0.33)

= 2.54 m/s <<<<<<<<<-----------------Answer

d) the momentum of mass m2 just after the elastic collision = change in momentum of m1

= 1*4.43 - 1*(-2.54)

= 6.97 kg.m/s <<<<<<<<<-----------------Answer

e) the kinetic energy of mass m2 just after the elastic collision = loss of energy of m1

= 6.566 J   <<<<<<<<<-----------------Answer

f) KE/P = (1/2)*m*v^2/(m*v)

KE/P = (1/2)*v

v = 2*KE/P

= 2*6.566/6.97

= 1.88 m/s   <<<<<<<<<-----------------Answer

g)
we know, KE = P^2/(2*m2)

m2 = P^2/(2*KE)

= 6.97^2/(2*6.566)

= 3.7 <<<<<<<<<-----------------Answer