Question

Two ice skaters approach each other at right angles. Skater A has a mass of 27.4 kg and travels in the x direction at 1.23 m/s. Skater B has a mass of 49.4 kg and is moving in the y direction at 0.759 m/s. They collide and cling together. Find the final speed of the couple. Answer in units of m/s.

Answer 1

The final speed will be "0.65 m/s".

Step-by-step explanation:

The given values are:

`m_a` = 27.4 kg

`m_b` = 49.4 kg

`v_a` = 1.23 m/s

`v_b` = 0.759 m/s

By using the law of conservation of momentum, we get

⇒
`P=√(P_1^2+P_2^2)`

⇒
`= √((m_av_a)^2+(m_bv_b)^2)`

On substituting the values, we get

⇒
`=√((27.4* 1.23)^2+(49.4* 0.759)^2)`

⇒
`=√(1135.82+1405.84)`

⇒
`=50.41 \ kg-m/s`

As we know,

⇒
`P=(m_1+m_2)v`

Then,

⇒
`v=(P)/(m_1+m_2)`

On substituting the values, we get

⇒
`=(50.41)/(27.4+49.4)`

⇒
`=(50.41)/(76.8)`

⇒
`=0.65 \ m/s`