Question

A skateboarder with a mass of 53-kg runs east with a speed of 4. 5 m/s toward her 3. 0-kg skateboard that is moving 6. 0 m/s west, and jumps on. What is the final speed and direction of the skateboarder on the board?.

Answer 1

I got 4.58m/s

Step-by-step explanation:

Using the conservation of momentum formula;

Total momentum before = Total momentum after

In this case, the equation is like so;

(mₐvₐ) + (mₙvₙ) = mₐₙvₓ

This equation shows that the skateboarder and their skateboard will move together as "one object" when the skateboarder jumps onto their skateboard! "vₓ" will represent the final velocity that we're trying to find out.

So;

mₐ = 53kg

vₐ = 4.5m/s

mₙ = 3

vₙ = 6

Substitute the values;

(53 × 4.5) + (3 × 6) = (53 + 3) × vₓ

238.5 + 18 = 56vₓ

256.5 = 56vₓ

vₓ = 256.5/56

∴ vₓ = 4.58m/s, to the east

The skateboarder had a greater initial momentum running from east. Hence, the direction will be towards the east!

I hope this helps! Let me know if I have any misconceptions or miscalculations! :)