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

Momentum of 2nd ball is

`P = 31.6 kg m/s`

direction is given as

`\theta = -37.66 degree`

Step-by-step 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 = √(P_2x)^2 + P_(2y)^2}`

`P = 31.6 kg m/s`

direction is given as

`tan\theta = (P_(2y))/(P_(2x))`

`tan\theta = (-19.3)/(25)`

`\theta = -37.66 degree`