Question

Ball 1 travels with a momentum of 48.0 kg-m/s east and strikes Ball 2, which is initially at rest.. Ball 1 separates at an angle of θ = 40.0º and a momentum of 30.0 kg-m/s. What is the magnitude and direction of the momentum of Ball 2 after the collision?

Answer 1

Answer:

Momentum of 2nd ball is

$P = 31.6 kg m/s$

direction is given as

$\theta = -37.66 degree$

Explanation:

As we know that there is no external force on the system of balls so momentum before and after collision will be conserved

So we have

$P_i = 48 \hat i + 0$

now after collision momentum of two balls is must be same as initial

so we have

$P_i = P_f$

$48\hat i = (30 cos40 \hat i + 30 sin40\hat j) + (P_{2x}\hat i + P_{2y}\hat j)$

so we have

$48 = 23 + P_{2x}$

$P_{2x} = 25 kg m/s$

for other component we have

$0 = 19.3 + P_{2y}$

$P_{2y} = -19.3 kg m/s$

Momentum of 2nd ball is given as

$P = \sqrt{P_2x}^2 + P_{2y}^2}$

$P = 31.6 kg m/s$

direction is given as

$tan\theta = \frac{P_{2y}}{P_{2x}}$

$tan\theta = \frac{-19.3}{25}$

$\theta = -37.66 degree$