Question

wo skaters collide and embrace in an inelastic collision. Alex's mass is 90 kg and his initial velocity is 1.5 m/s i . Barbara's mass is 57 kg and her initial velocity is 2.0 m/s j. After the collision, the two skaters move together at a common velocity. Use conservation of momentum to find their final speed. (Hint, find combined velocity vector first.)

Answer 1

1.21 m/s

Step-by-step explanation:

mass of Alex, mA = 90 kg

initial velocity of Alex, uA = 1.5 i m/s

mass of Barbara, mB = 57 kg

initial velocity of Barbara, uB = 2 j m/s

Let v is the velocity of combined mass after collision.

Use conservation of momentum

`90* 1.5\widehat{i}+57* 2\widehat{j}=(90+57)\overrightarrow{v}`

`\overrightarrow{v} = 0.92\widehat{i}+0.78\widehat{j}`

`v=\sqrt{0.92^(2)+0.78^(2)}`

v = 1.21 m/s

Thus, the velocity of combined mass is 1.21 m/s.

Answer 2

The two skaters will move with a common speed of 1.19 m/s.

Step-by-step explanation:

Given that,

Mass of Alex,
`m_1=90\ kg`

Initial velocity of Alex,
`u_1=1.5i\ m/s`

Mass of Barbara,
`m_2=57\ kg`

Initial velocity of Barbara,
`u_2=2j\ m/s`

After the collision, the two skaters move together at a common velocity. Let V is the common velocity. Using the conservation of momentum as :

`m_1u_1+m_2u_2=(m_1+m_2)V\\\\90* 1.5i+57* 2j=(90+57)V\\\\V=(135i+114j)/(147)\\\\V=(135i)/(147)+(114j)/(147)\\\\V=(0.91i+0.77j)\ m/s`

Magnitude of final velocity:

`|V|=√(0.91^2+0.77^2) \\\\|V|=1.19\ m/s`

So, the two skaters will move with a common speed of 1.19 m/s.