Question

A ball of mass 0.5kg moving horizontally with a velocity of 30 m/s strikes a vertical wall and rebounds horizontally with a velocity of 20 m/s. Calculate the impulse (Ft) exerted by the wall on the ball and the fore acting if thee time of contact of the ball is 0.045s.

Answer 1

Answer on Question 49997, Physics, Mechanics | Kinematics | Dynamics

Question:

A ball of mass $0.5\mathrm{kg}$ moving horizontally with a velocity of $30~\mathrm{m/s}$ strikes a vertical wall and rebounds horizontally with a velocity of $20~\mathrm{m/s}$. Calculate the impulse (Ft) exerted by the wall on the ball and the force acting if the time of contact of the ball is 0.045s.

Solution:

Let us find the impulse exerted by the wall on the ball. By the definition of the impulse:

$$\Delta p = p _ {1} - p _ {2} = m _ {b} v _ {b} - \left(- m _ {b} v _ {b} ^ {\prime}\right) = 0.5\mathrm{kg} \cdot 30\frac{\mathrm{m}}{\mathrm{s}} - \left(-0.5\mathrm{kg} \cdot 20\frac{\mathrm{m}}{\mathrm{s}}\right) = 25\mathrm{N} \cdot \mathrm{s},$$

where $m_{b}$ is the mass of the ball, $v_{b}$ is the velocity of the ball before strike with the vertical wall, $v_{b}^{\prime}$ is the velocity of the ball after rebound from the wall and we choose it with sign minus as it directed opposite to the $v_{b}$.

In order to obtain the force acting on the ball we again use the definition of the impulse:

$$\bar {F} \Delta t = \Delta p,$$

from this formula we can obtain $\bar{F}$:

$$\bar {F} = \frac {\Delta p}{\Delta t} = \frac {25\mathrm{N} \cdot \mathrm{s}}{0.045\mathrm{s}} = 555.5\mathrm{N}.$$

Answer:

1) $\Delta p = 25\mathrm{N} \cdot \mathrm{s}$.

2) $\bar{F} = 555.5\mathrm{N}$.

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