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

Answer:

Explanation:

solution

Answer 2

Answer

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 = $\dfrac{v_{2f} - v_{1f}}{v_{1i} - v_{2i})}$

0.73 = $\dfrac{v_{2f} - v_{1f}}{1-(-5))}$

v_{2f} − v_{1f} = 4.38

on solving

$v_{1f} = -1.938\ m/s$

$v_{2f} = 2.442\ m/s$