Question

A 4kg ball having velocity (7i+6j) m/s collides and bounces off a wall with a velocity of (-3i+6j) m/s. The ball is in contact with the wall for 0.01 s.In unit-vector notation, what are a) the impulse and b) the average force on the ball from the wall?

Answer 1

Answer on Question#52865 - Physics - Mechanics - Kinematics - Dynamics

A $m = 4\mathrm{kg}$ ball having velocity $\pmb{v}_{i} = (7\pmb{i} + 6\pmb{j})\frac{\mathrm{m}}{\mathrm{s}}$ collides and bounces off a wall with a velocity of $\pmb{v}_{f} = (-3\pmb{i} + 6\pmb{j})\frac{\mathrm{m}}{\mathrm{s}}$ . The ball is in contact with the wall for $\Delta t = 0.01\mathrm{s}$ . In unit-vector notation, what are a) the impulse and b) the average force on the ball from the wall?

Solution:

The change in impulse is given by

$$\Delta \boldsymbol {p} = m \left(\boldsymbol {v} _ {f} - \boldsymbol {v} _ {i}\right) = 4 \mathrm {k g} \cdot \left(- 3 \boldsymbol {i} + 6 \boldsymbol {j} - (7 \boldsymbol {i} + 6 \boldsymbol {j})\right) \frac {\mathrm {m}}{\mathrm {s}} = (- 4 0 \boldsymbol {i} + 0 \boldsymbol {j}) \frac {\mathrm {k g} \cdot \mathrm {m}}{\mathrm {s}}$$

The average force on the ball from wall could be found from the following relation

$$\boldsymbol {F} _ {a v} \cdot \Delta t = \Delta \boldsymbol {p}$$

Therefore,

$$\boldsymbol {F} _ {a v} = \frac {\Delta \boldsymbol {p}}{\Delta t} = \frac {\left(- 4 0 \boldsymbol {i} + 0 \boldsymbol {j}\right) \frac {\mathrm {k g} \cdot \mathrm {m}}{\mathrm {s}}}{0 . 0 1 \mathrm {s}} = \left(- 4 \boldsymbol {i} + 0 \boldsymbol {j}\right) \mathrm {k N}$$

Answer:

a) $(-40i + 0j)\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{s}}$

b) $(-4i + 0j)\mathrm{kN}$

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