Question

Two ice skaters approach each other at right angles. Skater A has a mass of 78.8 kg and travels in the +x direction at 3.16 m/s. Skater B has a mass of 29.9 kg and is moving in the +y direction at 2.01 m/s. They collide and cling together. Find the final speed of the couple. Answer in units of m/s.

Answer 1

Answer:

V = 22.35 m/s

Explanation:

Given that,

Mass of skater A, m = 78.8 kg

Speed of skater A, v = 3.16 m/s (+x direction)

Mass of skater B, m' = 29.9 kg

Speed of skater B, v' = 2.01 m/s (+y direction)

To find,

The final speed of the couple

Solution,

As both skaters collide and cling together. It is the case of inelastic collision in which the momentum of the system remains conserved,

Momentum of skater A, $p=mv=78.8\times 3.16=249.008i\ kgm/s$

Momentum of skater B, $p'=m'v'=29.9\times 2.01=60.099j\ kgm/s$

Total initial momentum,

$p_i=\sqrt{p^2+p'^2}$

$p_i=\sqrt{249.008^2+60.099^2}$

$p_i=256.15\ kgm/s$

According to the law of conservation of momentum,

$p_i=p_f$

Final momentum is equal to,

$p_f=(m+m')V$

$p_f=(78.8+29.9)V$

$256.15=(78.8+29.9)V$

V = 22.35 m/s

Therefore, the final speed of the couple is 22.35 m/s.