Question

Question #1bdce

Answer 1

J = 0.12

F = 90

See below for units and details :)

Explanation:

Maybe, the key to this question is spotting and understanding the symmetry which follows from this statement: "assume that the acceleration of the ball is constant while the ball is in contact with the ball" .

This means that the collision is elastic .

To demonstrate ( regrettably there is no native Socratic drawing tool to help), if we imagine the ball is thrown left to right to strike the ball, with positive velocity, `mathbf v`.

In that case, the acceleration vector, `mathbf a`, will point right to left, and it is of constant magnitude. If the ball goes from `15 ms^(-1)` to `0 ms^(-1)` over the `0.5 \ cm` distance, it must attain a velocity of `- 15 \ ms^(-1)` as it leaves contact with the wall.

The equation of motion, `v^2 = u^2 + 2 as`, which is a scalar energy construct, might nonetheless help make this clear.

Part (a)

Impulse `mathbf J` for constant force over a time period `Delta t` is defined as `mathbf J = mathbf F Delta t`.

With Newton's 2nd Law, and factoring in the constant acceleration, `mathbf F = m mathbf a = m (Delta mathbf v)/(Delta t)`, we reach:

`mathbf J = m Delta mathbf v`.

So the Impulse, `mathbf J`, is:

`mathbf J = 4cdot 10^(-3) times ((-15) - 15) = -0.12 \ Ns`.

This vector points to the left; and its magnitude is: `|mathbf J| = 0.12 \ Ns`.

Part (b)

To determine the force between wall and ball, we can use the scalar equation of motion:

`v^2 = u^2 + 2 as`

Over the period in which the ball is moving left to right, we say that:

`0^2 = 15^2 + 2 a(0.5 times 10^(-2))`

`implies a = - 22, 500 \ ms^(-2)`

This is acceleration that acts right to left. And because acceleration is constant, so is the force that is acting between the wall and ball.

Thanks to Newton's 2nd Law, and because `abs( mathbf a ) = 22, 500 \ ms^(-2)`, then:

`abs( mathbf F) = m abs( mathbf a) = 4 times 10^(-3) cdot 22, 500 \ ms^(-2) = 90 \ N`