Question

A 0.060-kg tennis ball, moving with a speed of 5.50m/s , has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.40m/s . Assume that the collision is perfectly elastic.
A. Determine the speed of the 0.060-kg ball after the collision.
B. Determine the direction of the velocity of the 0.060-kg ball after the collision.
Determine the direction of the velocity of the 0.060- ball after the collision.
1. in the direction of the initial velocity
2. in the direction opposite to the initial velocity
C. Determine the speed of the 0.090-kg ball after the collision.
D. Determine the speed of the 0.090-kg ball after the collision.
Determine the direction of the velocity of the 0.090- ball after the collision.
1. in the direction of the initial velocity
2. in the direction opposite to the initial velocity

Answer 1

Answer:

Explanation:

For elastic collision , the formula for velocity of .06 kg  is

v₁ = (m₁-m₂)/(m₁+m₂) u₁ + 2m₁m₂/(m₁+m₂) u₂

=( .06-.09 / .06+.09 ) 5.5 + (2 x .06 x .09 / .06+.09 ) 3.4

=( -.03 / .15) x 5.5 + (2 x .0054 / .15) x 3.4

= -1.1 +.2448

= - 0 .8552 m/s

Its direction will be - opposite direction

the formula for velocity of .09kg  is

v₂ = (m₂-m₁)/(m₁+m₂) u₂ + 2m₁m₂/(m₁+m₂) u₁

= ( .09-.06 / .06+.09 ) 3.4 + (2 x .06 x .09 / .06+.09 ) 5.5

= .68 +.396

= 1.076 m / s

Its direction will be in the same direction .