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A particle of mass m collides with anoth...

A particle of mass `m` collides with another stationary particle of mass `M` such that the second particle starts moving and the first particle stops just after the collision. Then which of the following conditions must always be valid ?

A

`m/Mle1`

B

`m/M=1`

C

`e=1`

D

`ele1`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the collision between two particles, one of mass \( m \) that is moving and another of mass \( M \) that is stationary. After the collision, the first particle comes to rest, and the second particle starts moving. We need to find the conditions that must always be valid after this collision. ### Step-by-Step Solution: 1. **Understanding the Initial Conditions**: - Let the mass \( m \) be moving with an initial velocity \( u \). - The mass \( M \) is stationary, so its initial velocity is \( 0 \). 2. **Applying the Conservation of Momentum**: - According to the law of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision. - Before the collision, the momentum is: \[ \text{Initial Momentum} = m \cdot u + M \cdot 0 = m \cdot u \] - After the collision, the first particle (mass \( m \)) stops, so its momentum is \( 0 \), and the second particle (mass \( M \)) moves with some velocity \( v \): \[ \text{Final Momentum} = m \cdot 0 + M \cdot v = M \cdot v \] - Setting initial momentum equal to final momentum gives us: \[ m \cdot u = M \cdot v \] 3. **Rearranging the Equation**: - From the equation \( m \cdot u = M \cdot v \), we can express the velocity of the second particle in terms of the first: \[ v = \frac{m}{M} \cdot u \] 4. **Understanding the Coefficient of Restitution**: - The coefficient of restitution \( e \) is defined as the ratio of the relative velocity of separation to the relative velocity of approach. - In this case, since the first particle comes to rest, the relative velocity of separation is \( v \) (the velocity of mass \( M \)) and the relative velocity of approach is \( u \) (the velocity of mass \( m \)): \[ e = \frac{v}{u} \] - Substituting for \( v \) from the previous step: \[ e = \frac{m}{M} \] 5. **Analyzing the Conditions**: - Since the coefficient of restitution \( e \) must always be less than or equal to 1 (as it represents a physical quantity related to energy loss during the collision), we have: \[ \frac{m}{M} \leq 1 \] - This implies that \( m \leq M \). ### Conclusion: The conditions that must always be valid after the collision are: 1. The coefficient of restitution \( e \) is less than or equal to 1. 2. The mass \( m \) must be less than or equal to mass \( M \). ### Final Answer: The correct options that must always be valid are: - \( e \leq 1 \) - \( m \leq M \)

To solve the problem, we need to analyze the collision between two particles, one of mass \( m \) that is moving and another of mass \( M \) that is stationary. After the collision, the first particle comes to rest, and the second particle starts moving. We need to find the conditions that must always be valid after this collision. ### Step-by-Step Solution: 1. **Understanding the Initial Conditions**: - Let the mass \( m \) be moving with an initial velocity \( u \). - The mass \( M \) is stationary, so its initial velocity is \( 0 \). ...
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