Answered: "A 0.50 kg football is thrown with a..."

College
Physics
Advanced Physics

Foemicofeduz

Answered question

2023-01-19

A 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.020 s. What is the force exerted on the receiver?

Answer & Explanation

peannreu

Beginner2023-01-20Added 10 answers

The following equation can be used to determine the average force applied:

F_{avg} = \frac{Δ p}{Δ t}

Where

Δ p

is the change in momentum and

Δ

is the change in time.
We are informed of the following:

\mapsto m = 0.50 kg

\mapsto v_{i} = 15 m / s  to the right

\mapsto v_{f} = 0

\mapsto Δ t = 0.020 s

Momentum is given by

\vec{p} = m \vec{v}

.
Therefore, the change in momentum

Δ p

is:

Δ p = p_{f} - p_{i} = m v_{f} - m v_{i}

We can calculate

Δ p

Δ p = (0.50 kg) (0 - 15 m / s)

\Rightarrow - 7.5 kgm / s

Keep in mind that the negative sign denotes a loss of momentum.

F_{avg} = \frac{- 7.5 kgm / s}{0.020 s}

= 375 N

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