Two smooth balls A and B, each of mass m and radius R, have their centres at (0, 0, R) and at (5R, – R, R) respectively, in a coordinate system as shown. Ball A, moving along positive x axis, collides with ball B. Just before the collision, speed of ball A is 4 m/s and ball B is stationary. The collision between the balls is elastic.
(i) Velocity of the ball A just after the collision is :
Text Solution
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(i) During collision, forces act along line of impact. As collision is elastic and both the balls have same mass, velocities are exchanged along the line of impact. Therefore ball B moves with velocity V B|| , that is equal to u cos 30°. Ball A moves perpendicular to the line of impact with velocity V A ⊥ = u cos60°. Along the line of impact, ball A does not have any velocity after the collision. Therefore velocity of ball A in vector form after the collision
= V A ⊥ cos60°i + V A ⊥ cos 30°j
= (u cos 60°) cos60°i + (u cos 60°) cos 30°j
= 
+ 
= 
m/s
(ii) Using impulse-momentum equation for ball B
and as 
= (mu cos 30°) cos 30 i – (mu cos30°) cos 60° j
= 
– 
= (3 m i – 
m j) kg 
(iii) Suppose V 2 is velocity of ball B along the line of impact and V 1 is velocity of ball A along the line of impact, after the collision, as shown.
Then 
(Velocity of approach) = Velocity of separation
= V 2 – V 1 .... (1)
Conserving momentum along the line of impact
m. u 
= m. V 2 + mV 1 .... (2)
Solving and using u = 4 m/s
V 2 = 
m/s
= 
m/s
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