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A body of mass m collides elastically wi...

A body of mass m collides elastically with another body at rest and then continues to move in the original continues to move in the original direction with one half of its original speed. mass of the body is

A

`m`

B

`2/3 m`

C

`m/3`

D

`m/2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the principles of conservation of momentum and the properties of elastic collisions. Here’s a step-by-step breakdown of the solution: ### Step 1: Define the Variables Let: - Mass of the first body = \( m \) - Initial velocity of the first body = \( V \) - Mass of the second body = \( M \) (unknown) - Initial velocity of the second body = \( 0 \) (at rest) - Final velocity of the first body after collision = \( \frac{V}{2} \) - Final velocity of the second body after collision = \( V_2 \) (unknown) ### Step 2: Apply Conservation of Momentum According to the law of conservation of momentum: \[ \text{Initial Momentum} = \text{Final Momentum} \] The initial momentum is: \[ mV + 0 = mV \] The final momentum is: \[ m \left(\frac{V}{2}\right) + MV_2 \] Setting the initial momentum equal to the final momentum gives us: \[ mV = m \left(\frac{V}{2}\right) + MV_2 \] ### Step 3: Simplify the Momentum Equation Rearranging the equation: \[ mV - m\left(\frac{V}{2}\right) = MV_2 \] This simplifies to: \[ mV - \frac{mV}{2} = MV_2 \] \[ \frac{mV}{2} = MV_2 \tag{1} \] ### Step 4: Use the Coefficient of Restitution For an elastic collision, the coefficient of restitution \( e = 1 \). The formula for the coefficient of restitution is: \[ e = \frac{\text{Relative velocity after collision}}{\text{Relative velocity before collision}} \] Thus: \[ V_2 - \frac{V}{2} = 1 \cdot \left(V - 0\right) \] This simplifies to: \[ V_2 - \frac{V}{2} = V \] Rearranging gives: \[ V_2 = V + \frac{V}{2} = \frac{3V}{2} \tag{2} \] ### Step 5: Substitute \( V_2 \) Back into the Momentum Equation Substituting equation (2) into equation (1): \[ \frac{mV}{2} = M \left(\frac{3V}{2}\right) \] Cancelling \( V \) from both sides (assuming \( V \neq 0 \)): \[ \frac{m}{2} = \frac{3M}{2} \] Multiplying both sides by 2: \[ m = 3M \] Thus, we can express \( M \) in terms of \( m \): \[ M = \frac{m}{3} \] ### Conclusion The mass of the second body is: \[ \boxed{\frac{m}{3}} \]
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Knowledge Check

  • A body of mass m kg collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body ?

    A
    m kg
    B
    2/3 m kg
    C
    m/3 kg
    D
    m/2 kg
  • A body of mass 3 kg collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body?

    A
    `1kg`
    B
    `1.5kg`
    C
    `2kg`
    D
    `5kg`
  • A body of mass 2 kg makes an elastic head - on collision another body at rest and continues to move in the original direction with one fourth of its original speed . The mass of the second body which collides with the first body is

    A
    `2 kg`
    B
    `1.2 kg`
    C
    `3 kg`
    D
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