Question

A force in the negative direction of an x-axis is applied for 27 msto a 0.40 kgball initially moving at14 m/sin the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude32.4 N.s. What are the balls (a) Speed and (b) Direction of travel just after the force is applied? What are (c) The average magnitude of the force and (d) The direction of the impulse on the ball?

Short answer

1. The speed of the ball just after the force was applied,$v_f=67\mathrm{m}/\mathrm{s}$
2. The direction of travel is along the negative x-axis.
3. The average magnitude of the force,${\mathrm{F}}_{\mathrm{avg}}=1.2\times {10}^3\mathrm{N}$
4. The impulse direction is along the negative x-axis.

Step-by-step solution

Understanding the given information

The mass of a ball,$\mathrm{m}=0.40\mathrm{kg}$.

The height of the victim from the floor,$\mathrm{h}=0.50\mathrm{m}$.

The time for which force is applied,$\mathrm{t}=27\mathrm{ms}\left(\frac{1\times {10}^{-3}\mathrm{s}}{1\mathrm{ms}}\right)=2.7\times {10}^{-2}\mathrm{s}$.

The initial velocity of a ball,${\mathrm{v}}_1=14\mathrm{m}/\mathrm{s}$.

The magnitude of the impulse, $\mathrm{J}=32.4\mathrm{N}·\mathrm{s}$.

Concept and Formula used in the given question

We can use the equation of impulse related to the change in momentum to calculate the speed of the ball after the impact. The magnitude of the average force can be calculated using an equation relating impulse, force, and time and equations are given as.

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ ∆\mathrm{P}=\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}\times \mathrm{t} \end{gathered}$$

(a) Calculation for the speed

The equation of impulse-momentum theorem as,

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ \mathrm{J}=\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ -32.4\mathrm{N}·\mathrm{s}=0.40\mathrm{kg}\times \left({\mathrm{v}}_{\mathrm{f}}-14\mathrm{m}/\mathrm{s}\right) \\ {\mathrm{v}}_{\mathrm{f}}-14\mathrm{m}/\mathrm{s}=\frac{-32.4\mathrm{N}.\mathrm{s}}{0.40\mathrm{kg}} \\ {\mathrm{v}}_{\mathrm{f}}-14\mathrm{m}/\mathrm{s}=-81\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{f}}=-67\mathrm{m}/\mathrm{s} \end{gathered}$$

Since the impulse is negative that means it is in the negative x-direction.

So, the speed of the ball is,

${\mathrm{v}}_{\mathrm{f}}=67\mathrm{m}/\mathrm{s}$

(b) Calculation for the Direction of travel just after the force is applied

The negative sign of $v_f$ indicates that the direction of travel is opposite of the initial direction. So, the direction is along the negative x-axis.

(c) Calculation for the average magnitude of the force

The equation for average force is

$$\begin{gathered} {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{\mathrm{t}} \\ =\frac{32.4\mathrm{N}·\mathrm{s}}{2.7\times {10}^{-2}\mathrm{s}} \\ =1.2\times {10}^2\mathrm{N} \end{gathered}$$

(d) Calculation for the direction of the impulse on the ball

The direction of impulse isthesame as the direction of applied force.

So, impulse direction is along the negative x-axis.