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

Answer:

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

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

⇒ $P=\sqrt{P_1^2+P_2^2}$

⇒     $= \sqrt{(m_av_a)^2+(m_bv_b)^2}$

On substituting the values, we get

⇒     $=\sqrt{(27.4\times 1.23)^2+(49.4\times 0.759)^2}$

⇒     $=\sqrt{1135.82+1405.84}$

⇒     $=50.41 \ kg-m/s$

As we know,

⇒ $P=(m_1+m_2)v$

Then,

⇒ $v=\frac{P}{m_1+m_2}$

On substituting the values, we get

⇒    $=\frac{50.41}{27.4+49.4}$

⇒    $=\frac{50.41}{76.8}$

⇒    $=0.65 \ m/s$