PHYSICS MIDDLE SCHOOL

A skateboarder traveling at 4.45 m/s can be stopped by a strong force in 1.82 s and by a weak force in 5.34 s.The mass of the skateboarder and the skateboard is 53.6 kg.

What is the impulse on the skateboarder?
What is the average force on the skateboarder in each of these stops?

Answers

Answer 1
Answer:

1) Impulse: -238.5 kg m/s

2) Strong force: -131.1 N, weak force: -44.7 N

Explanation:

1)

The impulse exerted on an object is equal to the change in momentum of the object itself.

Mathematically:

I=\Delta p=m(v-u)

where

m is the mass of the object

u is the initial velocity

v is the final velocity

For the skateboarder in this problem, we have:

m = 53.6 kg

u = 4.45 m/s

v = 0 (it comes to a stop)

Therefore, the impulse is

I=(53.6)(0-4.45)=-238.5 kg m/s

Where the negative sign indicates that the direction is opposite to the motion of the object.

2)

The impulse is also equal to the product between the force applied and the duration of the collision:

I=F\Delta t

where

I is the impulse

F is the average force

\Delta t is the time during which the force is applied

The strong force is applied in a time of

\Delta t = 1.82 s

Therefore this force is

F=(I)/(\Delta t)=(-238.5)/(1.82)=-131.1 N

The  weak force is applied in a time of

\Delta t = 5.34 s

So this force is

F=(I)/(\Delta t)=(-238.5)/(5.34)=-44.7 N

Learn more about impulse:

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