Question:

Two billiard balls of equal mass 30 g strike a rigid wall with same speed of 108 kmph (as shown) but at different angles. If the balls get reflected with the same speed then the ratio of the magnitude of impulses imparted to ball 'a' and ball 'b' by the wall along 'X' direction is : 

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The impulse imparted by a wall is always directed along the normal to the surface. To find the ratio, simply compare the normal components of the incident velocity: \(\frac{\cos \theta_1}{\cos \theta_2}\).
Updated On: Jan 19, 2026
  • $1 : \sqrt{2}$
  • $1 : 1$
  • $2 : 1$
  • $\sqrt{2} : 1$
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Impulse is defined as the change in momentum of a body.
When a ball reflects from a wall, only the component of velocity perpendicular to the wall changes direction.
Step 2: Key Formula or Approach:
Impulse (\(J\)) = Change in Momentum (\(\Delta p\)).
Along the X-direction (normal to the wall): \(J_x = m(v_x - u_x)\).
For elastic reflection: \(J_x = 2mu \cos \theta\), where \(\theta\) is the angle with the normal.
Step 3: Detailed Explanation:
Speed \(u = 108 \text{ kmph} = 108 \times \frac{5}{18} = 30 \text{ m/s}\).
Ball 'a' strikes normally: Angle with normal \(\theta_a = 0^\circ\).
\[ J_a = 2mu \cos 0^\circ = 2mu \]
Ball 'b' strikes at an angle: From the figure, the angle with the wall is \(45^\circ\), so the angle with the normal is also \(\theta_b = 45^\circ\).
\[ J_b = 2mu \cos 45^\circ = 2mu \left(\frac{1}{\sqrt{2}}\right) = \sqrt{2}mu \]
The ratio of impulses is:
\[ \frac{J_a}{J_b} = \frac{2mu}{\sqrt{2}mu} = \frac{2}{\sqrt{2}} = \frac{\sqrt{2}}{1} \]
Step 4: Final Answer:
The ratio of magnitudes of impulses imparted along the X direction is \(\sqrt{2} : 1\).
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