Question

You are lawn bowling and bowl a 350 g ball with a speed of 5.94 m/s. Unfortunately, your ball hits a large rock, resulting in a perfectly elastic collision. If the final velocity of the rock after the collision is 3.69 m/s, what must the mass of the rock be?

* 0.350 kg
* 0.0683 kg
* 0.777 kg
* 0.160 kg

****I tried this using conservation of kinetic energy because it says the collision is perfectly elastic, and ended up with 0.907 kg as my answer. Not sure what I'm doing wrong.****

Answer 1

here the velocity of the ball is missing after collision

like m1u1 + m2u2 = m1v1 + m2v2

or, m1u1 = m1v1 + m2v2 as first the rock is at rest so, u2=0

so, 0.350x5.94 = 0.350 xv1 + m2 x3.69

so, v1 = (2.079 -3.69m2)/0.350

as elastic collision so kinetic energy remain fixed

so, 0.5m1u1^2 + 0.5m2u2^2 = 0.5m1v1^ + 0.5m2v2^2

or, m1u1^2 = m1v1^ + m2v2^2

or, 0.350x5.94^2 = 0.350 x(2.079- 3.69m2)^2/0.350^2 + m2 x3.69^2 substituiting value of v1

or, 12.34926 x0.350 = (2.079- 3.69m2)^2 + 4.7656m2

or, 4.3222 = 4.3222 -15.343m2 + 13.6161 (m2)^2 + 4.7656m2

or, 13.6161(m2)^2 -10.5774m2 = 0

or, m2 (13.6161m2 - 10.5774)

or, we can write (13.6161m2 - 10.5774) = 0

or, m2 = 0.7768 =0.777 kg......................................ans