Question

# Two blocks of masses m1 = 1.50 kg and m2 = 3.00 kg are each released...

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.

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

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