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Question
A ball of mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence 45°. The ball is reflected by the plane at equal angle of reflection with the same speed. Calculate (a) the magnitude of the change in momentum of the ball (b) the change in the magnitude of the momentum of the ball.
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Solution
It is given that:
Mass of the ball = 50 g =
\[0 . 05 \text{ Kg}\]
Speed of the ball, v = 2.0 m/s
Incident angle = 45˚
\[v = 2 \cos 45^\circ \hat i - 2 \sin 45^\circ \hat j \]
\[v' = - 2 \cos 45^\circ \hat i - 2 \sin 45^\circ \hat j\]
\[(a) \text{ Change in momentum } = m \vec{v} - m \vec{v} '\]
\[ = 0 . 05\left( 2 \cos 45^\circ\ \hat i - 2 \sin 45^\circ \hat j \right) - 0 . 05\left( - 2 \cos 45^\circ \hat i - 2 \sin 45^\circ \hat j \right)\]
\[ = 0 . 1 \cos 45^\circ\ \hat i - 0 . 1 \sin 45^\circ \ \hat j + 0 . 1 \cos 45^\circ \ \hat i + 0 . 1 \sin 45^\circ\ \hat j\]
\[ = 0 . 2 \cos 45^\circ\ \hat i \]
\[ \therefore \text{ Magnitude }= \frac{0 . 2}{\sqrt{2}} = 0 . 14 \text{ Kg m/s}\]
(b) The change in magnitude of the momentum of the ball,
\[\left| \vec{P}_2 \right| - \left| \vec{P}_1 \right| = 2 \times 0 . 5 - 2 \times 0 . 5 = 0\]
i.e. There is no change in magnitude of the momentum of the ball.
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