Question

A .060kg tennis ball, moving with a speed of 4.5m/s has a head-on collision with a .090kg clay ball initially moving in the same direction at a speed of 3.9 m/s. Assuming perfect elastic collision, determine the speed and direction of the tennis ball

Answer 1

Given that,

m1 = 0.06 kg

v1i = 4.5 m/s

m2 = 0.09 kg

v2i = 3.9 m/s

From conservation of the momentum,

m1v1i + m2v2i = m1v1f + m2v2f  

v2f = (m1v1i + m2v2i - m1v1f) / m2 ..........(1)

From conservation of the energy,

(1/2)m1v1i² + (1/2)m2v2i² = (1/2)m1v1f² + (1/2)m2v2f²

m1v1i² + m2v2i² = m1v1f² + m2v2f²

Put the value of v2f from eq (1),

m1v1i² + m2v2i² = m1v1f² + m2((m1v1i + m2v2i - m1v1f) / m2 )²

m1v1i² + m2v2i² = m1v1f^2 + (m1v1i + m2v2i)²/m2 + 2m1*v1f(m1v1i + m2v2i)/m2 + m1²v1f² / m2

[(m1 + m1²/m2) v1f²] - [(2m1*(m1v1i + m2v2i) / m2) v1f] + [(m1v1i + m2v2i)²/m2 - m1v1i² + m2v2i²] = 0

By putting and solving all the values,

(m1 + m1²/m2) = 0.1

2m1*(m1v1i + m2v2i) / m2 = 0.828

[(m1v1i + m2v2i)²/m2 - m1v1i² + m2v2i²] = 1.701

0.1v1f² - 0.828v1f + 1.701 = 0

By solving above quadratic equation,

v1f = 3.78 m/s

Speed of the tennis ball = 3.78 m/s

Direction is same as before collision.