Question

A particle of mass \( m \) is projected with velocity \( v \) at an angle \( \theta \) with the horizontal. At its highest point, it explodes into two pieces of equal mass. One of the pieces continues to move on the original trajectory. The velocity of the second piece is:

• A. \( 2v \cos \theta \)
• B. \( v \cos \theta \)
• C. \( 3v \cos \theta \)
• D. \( \frac{v}{2} \cos \theta \)

Hint:

- At the highest point, only the horizontal velocity component remains. - Use momentum conservation along the horizontal direction. - Since one fragment continues with the same velocity, the second must also move with \( v \cos \theta \) to satisfy momentum conservation.

Answer 1

The Correct Option is B

Step 1: Velocity Components at the Highest Point At the highest point of the projectile's motion: - The vertical component of velocity becomes zero. - The horizontal component remains: \[ v_x = v \cos \theta \] Since there are no external horizontal forces, the total momentum in the horizontal direction is conserved. 
Step 2: Applying Conservation of Momentum Before the explosion: \[ P_{{initial}} = m v \cos \theta \] After the explosion: - The projectile splits into two equal parts of mass \( m/2 \). - One of the fragments continues with the same velocity \( v \cos \theta \). - Let the velocity of the second fragment be \( v_2 \). Applying the law of conservation of momentum in the horizontal direction: \[ m v \cos \theta = \frac{m}{2} v \cos \theta + \frac{m}{2} v_2 \] 
Step 3: Solving for \( v_2 \) \[ \frac{m}{2} v_2 = m v \cos \theta - \frac{m}{2} v \cos \theta \] \[ = \frac{m}{2} v \cos \theta \] \[ v_2 = v \cos \theta \] Thus, the velocity of the second fragment is: \[ v_2 = v \cos \theta \]