# Impulse and average force

#### Knfoster

1. Homework Statement

A 2.18 kg steel ball strikes a massive wall at 12.6 m/s at an angle of a = 55.1o with the plane of the wall. It bounces off with the same speed and angle (as seen in the figure below). If the ball is in contact with the wall for 0.152 s, what is the average force exerted on the ball by the wall?

2. Homework Equations

mv/change in t = average force
v=sq. root 2gh

3. The Attempt at a Solution
I've tried finding the impulse and then dividing it by the .152 sec. but I keep getting the wrong answer. Could someone please walk me through the equations and steps? Thanks!

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#### LowlyPion

Homework Helper
1. Homework Statement

A 2.18 kg steel ball strikes a massive wall at 12.6 m/s at an angle of a = 55.1o with the plane of the wall. It bounces off with the same speed and angle (as seen in the figure below). If the ball is in contact with the wall for 0.152 s, what is the average force exerted on the ball by the wall?

2. Homework Equations

mv/change in t = average force
v=sq. root 2gh

3. The Attempt at a Solution
I've tried finding the impulse and then dividing it by the .152 sec. but I keep getting the wrong answer. Could someone please walk me through the equations and steps? Thanks!
What is the Δ in momentum?

#### Knfoster

momentum= mass(velocity)...2.18*12.6=27.5

#### LowlyPion

Homework Helper
momentum= mass(velocity)...2.18*12.6=27.5
But what was the Δ in momentum.

Think x,y components.

#### sArGe99

Resolve the initial and final velocity vectors to find out which component undergoes a change.

#### Knfoster

v1 prime would be -v1
v2 prime would be 0

how does that fit in to the problem?

#### LowlyPion

Homework Helper
v1 prime would be -v1
v2 prime would be 0

how does that fit in to the problem?
Impulse requires knowing the Δ in momentum.

Now not all the momentum changes here. For instance the momentum || to the wall is unchanged after contacting the wall.

But the ⊥ component has a change. Figure that change as I think you will be needing it.

#### Knfoster

I don't understand... Is that not just the negative of the velocity?

#### LowlyPion

Homework Helper
I don't understand... Is that not just the negative of the velocity?
Negative of the original velocity means that it returned in the direction it came.

It did not.

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