A ball of mass 0.612 kg moving east (+ x direction) with a speed of 3.90 m/s col

Question

A ball of mass 0.612 kg moving east (+xdirection) with a speed of 3.90 m/s collides head-on with a 0.306 kg ball at rest. Assume that the collision is perfectly elastic.

Part A

What is be the speed of the 0.612-kg ball after the collision?

Part B

What is be the direction of the velocity of the 0.612-kg ball after the collision?

to the east or to the west?

Part C

What is the speed of the 0.306-kg ball after the collision?

Part D

What is the direction of the velocity of 0.306-kg ball after the collision?

What is the direction of the velocity of 0.306- ball after the collision?

Explanation / Answer

Given that,

m1 = 0.612 kg ; u1 = 3.90 m/s

m2 = 0.306 ; u2 = 0 (as it is on rest intially);

a) Let v1 be the velocity of m1 = 0.612 kg after the collision.

We know that, momentum and energy both are conserved in the elastic collisions.

P(i) = P(f)

m1u1 + m1u2 = m1v1 + m2v2

E(i) = E(f)

1/2 m1 u12 = 1/2 m1v12 + 1/2 m2 v22

from above two equations we get,

v1 = (m1 - m2)v1 / (m1 + m2)

v1 = (0.612 - 0.306) x 3.9 / (0.612 + 0.306) = 1.1934 / 0.918 = 1.3 m/s

Hence, v1 = 1.3 m/s

b)direction will remain East.

c)Let v2 be the speed me m2 = 0.306 kg

solving the equations in (a) for v2 we get:

v2 = 2 m1u1/( m1 + m2)

v2 = 2 x0.612 x 3.9 / (0.612 + 0.306) = 5.2 m/s

Hence, v2 = 5.2 m/s

d)direction of m2 will be towards east.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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