Question

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the adjacent Figure. It collides with an identical disk that is at rest before the collision. The collision is elastic. Disk 1 goes off at an 60 5.0 m/s angle of 60 with respect to the x-direction. Disk 2 g 30 goes off at an angle of 30 with respect to the x-direction. What best describes the speeds of the disks after the collision?

Answer 1

Let
`$m_1=m_2=4$` kg

`$u_1 = 5$` m/s

Let
`$v_1$` and
`$v_2$` are the speeds of the disk
`$m_1$` and
`$m_2$` after the collision.

So applying conservation of momentum in the y-direction,

`$0=m_1 .v_1_y -m_2 .v_2_y $`

`$v_1_y = v_2_y$`

`$v_1 . \sin 60=v_2. \sin 30$`

`$v_2 = v_1 * (\sin 60)/(\sin 30)$`

`$v_2=1.732 * v_1$`

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

`$m_1.u_1=m_1.v_1_x+m_2.v_2_x$`

`$u_1=v_1_x+v_2_x$`

`$5=v_1. \cos 60 + v_2 . \cos 30$`

`$5=v_1. \cos 60 + 1.732 * v_1 \cos 30$`

`$v_1 = 2.50$` m/s

So,
`$v_2 = 1.732 * 2.5$`

= 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s