Question

Two blocks of masses

m₁ = 1.50 kg

and

m₂ = 3.00 kg

are each released from rest at a height of

h = 4.40 m

on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)

(a) Determine the velocity of each block just before the collision.

(b) Determine the velocity of each block immediately after the collision.

(c) Determine the maximum heights to which

m₁

and

m₂

rise after the collision.

Answer 1

a) both masses are released from rest

by energy conservation

change in potential energy = change in kinetic energy

mgh = 1/2mv²

so v= sqrt(2gh) where sqrt = squar root

for mass1 m1= 1.5kg

V₁ = sqrt(2*9.8*4.40) = 9.29 m/sec

for mass M₂

V₂ = sqrt(2*9.8*4.40) = 9.29 m/sec

b) elastic collision both conservation of momentum and kinetic energy are observed

so velocity of M₁ after collision (V ₃) = 5*9.29/3 = 15.83 m/sec

and  velocity of M2 after collision (V ₄) = 9.29/3 = 3.1 m/sec

c)  again energy conservation

1/2 mv² = mgh

h = v² /2g

so mass 1 H₁ = 15.83² /2*9.8 = 12.78 m

and mass 2 H₂ = 3.1² / 2*9.8 = 0.49 meter