Question

Ques
A skateboarder is cruising on his skateboard at 3 m/s and has a total mass of 70 kg. A friend jumps on
(from standing still). What is the new velocity of the skateboarder if the friend has a mass of 50 kg?

Answer 1

Answer:

When a particle or a system of particles move in a system where no external force acts, then the total linear momentum of the particle system remains constant.

Explanation:

Given data:

Total mass of the skateboarder, $m_{1} = 70 \ kg$

Mass of the friend, $m_{2} = 50 \ kg$

Initial velocity of the skateboarder, $u_{1} = 3 \ m/s$

Initial velocity of the the friend, $u_{2} = 0$

Let the new velocity of the skateboarder when his friend jumps be $v$.

From the conservation law of linear momentum,

$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

$70 \times 3 + 50 \times 0 = (70 + 50)v$

$\Rightarrow \ v &= 1.75 \ m/s.$