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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

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प्रश्न

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.

योग
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उत्तर

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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Momentum Conservation and Centre of Mass Motion
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Centre of Mass, Linear Momentum, Collision - Exercise [पृष्ठ १६१]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 9 Centre of Mass, Linear Momentum, Collision
Exercise | Q 20 | पृष्ठ १६१

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