Question

Two ice skaters approach each other at right angles. Skater A has a mass of 32.7 kg and travels in the x direction at 2.3 m/s. Skater B has a mass of 93.3 kg and is moving in the y direction at 1.51 m/s. They collide and cling together. Find the final speed of the couple. Answer in units of m/s.

Answer 1

Answer:

Final speed, v = 1.25 m/s

Explanation:

Given that,

Mass of skater A, $m_A=32.7\ kg$

Initial speed of skater A, $u_A=2.3i\ m/s$ (x axis)

Mass of skater B, $m_B=93.3\ kg$

Initial speed of skater B, $u_B=1.51j\ m/s$ (y axis)

It is mentioned that the two skaters collide and cling together. It is case of inelastic collision in which momentum remains conserved. Let V is the final speed of the couple. It is given by :

$m_Au_A+m_Bu_B=(m_A+m_B)V$

$32.7\times 2.3i+93.3\times 1.51j=(32.7+93.3)V$

$V=\dfrac{75.21i+140.88j}{126}$

$V=(0.59i+1.11j)\ m/s$

$|V|=1.25\ m/s$

So, the final speed of the couple is 1.25 m/s. Hence, this is the required solution.