A 100 g ball moving to the right at 4.0 m/s catches up and collides with a 440 g

Question

A 100 g ball moving to the right at 4.0 m/s catches up and collides with a 440 g ball that is moving to the right at 1.0 m/s .

(A) If the collision is perfectly elastic, what is the speed of the 100 g ball after the collision?

(B) If the collision is perfectly elastic, what is the direction of motion of the 100 g ball after the collision?

(C) If the collision is perfectly elastic, what is the speed of the 440 g ball after the collision?

(D) If the collision is perfectly elastic, what is the direction of motion of the 440 g ball after the collision?

Explanation / Answer

A) Momentu will be conserved. So 0.1*4 + 0.44*1 = 0.1*V1 + 0.44*V2 = 0.84

Velocity of approach = -velocity of sepration. So V1 - V2 = 4 -1 = -3

=> V2 = V1 + 3 put this in above equation and find V1

V1 = -0.8889 m/s

So speed = 0.-0.8889 m/s

B) towards left

C ) V2 = V1+3 = 2.111 m/s

So speed = 2.111 m/s

D) Direction = towards right

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.