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WAEC Physics 2016 Paper

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Question : 9

Total: 47

At a birthday party, the celebrant pops a corked fruit wine. If the cork shoots out of the bottle at an angle of 40^o to the horizontal and travels a horizontal distance of 4.50m in 1.25s. Calculate the initial speed of the cork.

4.2ms^-1
4.7ms^-1
5.6ms^-1
7.1ms^-1

Solution:

Concept:

The horizontal motion of the cork is uniform, so horizontal distance

R

equals horizontal component of velocity

U\cos\theta

multiplied by time

t

.

Explanation:

Given:

R = 4.50\,\text{m}

,

\theta = 40^\circ

,

t = 1.25\,\text{s}

.

Using

R = U\cos\theta \times t

,

U = \dfrac{R}{\cos\theta \times t}

.

Substitute values:

U = \dfrac{4.50}{\cos 40^\circ \times 1.25}

.

U = \dfrac{4.50}{0.7660 \times 1.25}

.

U = \dfrac{4.50}{0.9575} \approx 4.7\,\text{m/s}

.

Thus the initial speed of the cork is approximately

4.7\,\text{m/s}

.

Answer:

Option B.

4.7\,\text{ms}^{-1}

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