Question

ball having a mass of 0.5 kg is moving towards the east with a speed of 8.0 ms-1.
After being hit by a bat it changes its direction and starts moving towards the north with
a speed of 6.0 ms−1. If the time of impact is 0.1 s, calculate the impulse and average
force acting on the ball.

Answer 1

Answer on Question 72586, Physics / Mechanics | Relativity

Question

Ball having a mass of $0.5\mathrm{kg}$ is moving towards the east with a speed of $8.0~\mathrm{ms - 1}$ .

After being hit by a bat it changes its direction and starts moving towards the north with

a speed of 6.0 ms-1. If the time of impact is 0.1 s, calculate the impulse and average

force acting on the ball.

Solution.

Momentum of a ball is

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

where $m$ is a mass of the ball, $\vec{v}$ is a velocity of the ball. Momentum $\vec{p}$ is a vector having the same direction as the velocity $\vec{v}$ .

Impulse is the change in momentum

$$\Delta \vec {p} = \vec {p} _ {f} - \vec {p} _ {i} = m \Delta \vec {v} = m (\vec {v} _ {f} - \vec {v} _ {i})$$

where subscripts i and f mean the initial and final values.

Let the direction to the east be the $x$ axis, and the direction to the north be the $y$ axis as shown in the figure

Since at first the ball moved towards the east, then $v_{i} = v_{ix} = 8.0\mathrm{m / s}$ and $v_{iy} = 0\mathrm{m / s}$ . After being hit by a bat it moved towards the north, then $v_{fx} = 0\mathrm{m / s}$ $v_{f} = v_{fy} = 6.0\mathrm{m / s}$ .

Then for $x$ -component of momentum we have

$$p _ {i} = p _ {i x} = m v _ {i x} = 0. 5 \mathrm {k g} \cdot 8. 0 \mathrm {m / s} = 4 \mathrm {k g} \cdot \mathrm {m / s}$$ $$p _ {f x} = m v _ {f x} = 0 \mathrm {k g} \cdot \mathrm {m / s}$$

for $y$ -component of momentum we have

$$p _ {i y} = m v _ {i y} = 0 \mathrm {k g} \cdot \mathrm {m / s}$$ $$p _ {f y} = m v _ {f y} = 0. 5 \mathrm {k g} \cdot 6. 0 \mathrm {m / s} = 3 \mathrm {k g} \cdot \mathrm {m / s}$$

The $x$ -component of impulse is

$$\Delta p _ {x} = p _ {2 x} - p _ {1 x} = 0 - 4 = - 4 \mathrm {k g} \cdot \mathrm {m / s}$$

The $y$ -component of impulse is

$$\Delta p _ {y} = p _ {2 y} - p _ {1 y} = 3 - 0 = 3 \mathrm {k g} \cdot \mathrm {m / s}$$

Magnitude of impulse is defined as

$$\Delta p = | \Delta \vec {p} | = \sqrt {\Delta p _ {x} ^ {2} + \Delta p _ {y} ^ {2}} = \sqrt {(- 4) ^ {2} + 3 ^ {2}} = \sqrt {2 5} = 5 \mathrm {k g} \cdot \mathrm {m / s}$$

Average force acting on the ball is defined as

$$\vec {\mathrm {F}} _ {a v} = \frac {\Delta \vec {\mathrm {p}}}{\Delta t}$$

Magnitude of average force acting on the ball is

$$\mathrm {F} _ {a v} = \left| \vec {\mathrm {F}} _ {a v} \right| = \frac {| \Delta \vec {\mathrm {p}} |}{\Delta t}$$

Substituting $|\Delta \vec{p}| = 5 \, \mathrm{kg} \cdot \mathrm{m/s}$ and the time of impact $\Delta t = 0.1 \, \mathrm{s}$ we get average force

$$\mathrm {F} _ {a v} = \frac {5 \, \mathrm{kg} \cdot \mathrm{m/s}}{0.1 \, \mathrm{s}} = 50 \, \mathrm{N}$$

Answer: impulse of the ball is $\Delta p = 5 \, \mathrm{kg} \cdot \mathrm{m/s}$ , average force acting on the ball is $\mathrm{F}_{av} = 50 \, \mathrm{N}$

Answer provided by AssignmentExpert.com