Question

One can write v1f − v2f = ǫ (v2i − v1i) where ǫ takes a value between 0 (perfectly inelastic) and 1 (perfectly elastic) for most collisions. Suppose two pucks sliding on frictionless ice collide head-on with a coefficient of restitution ǫ = 0.73. They have masses and initial velocities of m1 = 0.38 kg, m2 = 0.15 kg, v1i = 1 m/s, and v2i = −5 m/s. What is the velocity of puck 1 after the collision? Answer in units of m/s.

Answer 1

Step-by-step explanation:

solution

Answer 2

given,

ǫ = 0.73

m₁ = 0.38 kg m₂ = 0.15 kg

v₁i = 1 m/s, v₂i = −5 m/s

applying conservation equation of momentum

`m_1v_(1i) + m_2v_(2i) = m_1v_(1f) + m_2v_(2f)`

`(0.38)(1) + (0.15)(-5) = 0.38v_(1f) + 0.15v_(2f)`

`0.38v_(1f) + 0.15v_(2f) = -0.37`

v_{1f} − v_{2f} = ǫ (v_{2i} − v_{1i})

ǫ =
`(v_(2f) - v_(1f))/(v_(1i) - v_(2i)))`

0.73 =
`(v_(2f) - v_(1f))/(1-(-5)))`

v_{2f} − v_{1f} = 4.38

on solving

`v_(1f) = -1.938\ m/s`

`v_(2f) = 2.442\ m/s`