Question

A ten grams ball glides to the left at a speed of 0.4 meters per second on a frictionless flat surface at the OC fair skating rink. The ball collides head-on, elastically, with another larger ball with mass 30 grams which was gliding to the right with a speed of 0.2 meters per second.

A. After the collision, what is the final velocity of each ball.

B. What is the change in momentum for each ball? How do they compare to each other?

C. What is the change in kinetic energy for each ball? How do they compare to each other?

Answer 1

Solution) m1 = 10 grams

m2 = 30 grams

u1 = - 0.4 m/s (Left)

u2 = 0.2 m/s (right)

(A) Final velocity of each ball , V1 = ?

V2 = ?

V1 = [((m1 - m2)(u1))/(m1+m2)] + [(2(m2)(u2))/(m1+m2)]

V1 = [((10 - 30)(- 0.4))/(10+30)] + [(2×30×0.2)/(10+30)]

V1 = 0.5 m/s

V2 = [((2)(m1)(u1))/(m1+m2)] - [((m1 - m2)(u2))/(m1+m2)]

V2 = [(2×10×(-0.4))/(10+30)] - [((10 - 30)(0.2))/(10+30)]

V2 = - 0.2 + 0.1

V2 = - 0.1 m/s

(B) Change in momentum of ball one ,

P1 = (m1)(V1 - u1)

m1 = 10 g = 10×10^(-3) kg

m1 = 10^(-2) kg

P1 = 10^(-2)(0.5 - (-0.4))

P1 = 0.9×10^(-2) kgm/s

P1 = 9×10^(-3) kgm/s

Change in momentum of ball 2 ,

P2 = (m2)(V2 - u2)

P2 = (30×10^(-3))(-0.1 - 0.2)

P2 = 9×10^(-3) kgm/s

Therefore ,

P1 = P2

(C) change in kinetic energy of ball 1 ,

KE1 = (1/2)(m1)(V1^2 - u1^2)

KE1 = (1/2)(10×10^(-3))(0.5^2 - (-0.4)^2)

KE1 = 0.45×10^(-3) J

Change in Kinetic energy of ball 2 ,

KE2 = (1/2)(m2)(V2^2 - u2^2)

KE2 = (1/2)(30×10^(-3))((-0.1)^2 - (0.2)^2)

KE2 = 0.45×10^(-3) J

Therefore

KE1 = KE2