Question

A 0.060-kg tennis ball, moving with a speed of 5.40m/s , has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.38m/s . Assume that the collision is perfectly elastic.

Part A

Determine the speed of the 0.060-kg ball after the collision.

Part B

Determine the direction of the velocity of the 0.060- ball after the collision.

in the direction of the initial velocity

in the direction opposite to the initial velocity

Part C

Determine the speed of the 0.090-kg ball after the collision.

Part D

Determine the direction of the velocity of the 0.090- ball after the collision.

	in the direction of the initial velocity
	in the direction opposite to the initial velocity

Answer 1

according to conservation of linear momentum

m1*u1 + m2*u2 = m1*v1+m2*v2

m1*(u1-v1) = m2*(v2-u2)........(1)

according to conservation of energy

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

m1*(u1^2 - v1^2) = m2*(v2^2-u2^2).....(2)

from 1 &2

u1 + v1 = u2+v2

u1 - u2 = v2 - v1

v2 = u1 - u2 + v1........(3)

3 in 1

m1*(u1-v1) = m2*(u1 - u2 + v1 - u2)

v1 = u1*(m1-m2)/(m1+m2) + 2*m2*u2/(m1+m2)

v2 = u2*(m2-m1)/(m1+m2) + 2*m1*u1/(m1+m2)

m1 = 0.06 kg...........m2 = 0.09 kg

u1 = +5.4...........u2 = +3.38

v1 = ?...........v2 = ?

A) v1 = (((0.06-0.09)*5.4)/(0.06+0.09)) + ((2*0.09*3.38)/(0.06+0.09)) = + 2.976 m/s

B)     in the direction of the initial velocity

C) v2 = (((0.09-0.06)*3.38)/(0.06+0.09)) +
((2*0.06*5.4)/(0.06+0.09)) = +4.996 m/s

D) in the direction of the initial velocity