Home
Class 12
PHYSICS
A sphere of mass m falls on a smooth hem...

A sphere of mass m falls on a smooth hemisphere of mass M resting with its plane face on smooth horizontal table, so that at the moment of impact, line joining the centres makes an angle with the vertical. The velocity of sphere just before impact is u and e is the coefficient of restitution.

A

The velocity of hemisphere after impact will be zero

B

The hemisphere will move with a velocity `v = ( m u sin alpha)/(M cos^2 alpha)`

C

The hemisphere will move with a velocity `v = (m u cos alpha( 1 + e))/((M + m ))`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will analyze the situation using the principles of conservation of momentum and the coefficient of restitution. ### Step 1: Understand the Setup We have a sphere of mass \( m \) falling towards a smooth hemisphere of mass \( M \). The sphere is moving with a velocity \( u \) just before impact, and the line joining the centers of the sphere and hemisphere makes an angle \( \alpha \) with the vertical. ### Step 2: Resolve the Velocity of the Sphere Before the impact, we can resolve the velocity \( u \) of the sphere into two components: - The vertical component: \( u \cos \alpha \) - The horizontal component: \( u \sin \alpha \) However, since the hemisphere is smooth and can only move horizontally, we will focus on the horizontal component for momentum conservation. ### Step 3: Apply Conservation of Momentum According to the conservation of momentum, the momentum before the collision must equal the momentum after the collision. The momentum before the collision is given by the horizontal component of the sphere's velocity: \[ \text{Initial momentum} = m \cdot u \cos \alpha \] Let \( V \) be the velocity of the hemisphere after the impact and \( v \) be the velocity of the sphere after the impact. The momentum after the collision is: \[ \text{Final momentum} = M \cdot V + m \cdot v \] Setting the initial momentum equal to the final momentum: \[ m \cdot u \cos \alpha = M \cdot V + m \cdot v \tag{1} \] ### Step 4: Apply the Coefficient of Restitution The coefficient of restitution \( e \) relates the relative velocities of the two bodies before and after the collision. It is defined as: \[ e = \frac{\text{Relative velocity after collision}}{\text{Relative velocity before collision}} \] In our case, the relative velocity before the collision is \( u \cos \alpha \) (the sphere is moving towards the hemisphere). After the collision, the relative velocity is \( V + v \) (the hemisphere moves with velocity \( V \) and the sphere moves with velocity \( v \)): \[ e = \frac{V + v}{u \cos \alpha} \tag{2} \] From equation (2), we can express \( v \) in terms of \( V \): \[ v = e u \cos \alpha - V \tag{3} \] ### Step 5: Substitute Equation (3) into Equation (1) Now we substitute equation (3) into equation (1): \[ m \cdot u \cos \alpha = M \cdot V + m \cdot (e u \cos \alpha - V) \] Expanding this gives: \[ m \cdot u \cos \alpha = M \cdot V + m e u \cos \alpha - m V \] Rearranging terms: \[ m u \cos \alpha - m e u \cos \alpha = (M - m) V \] Factoring out \( u \cos \alpha \): \[ u \cos \alpha (m - m e) = (M - m) V \] ### Step 6: Solve for \( V \) Now we can solve for \( V \): \[ V = \frac{u \cos \alpha (m(1 - e))}{M + m} \] ### Step 7: Final Expression Thus, the final expression for the velocity of the hemisphere after the impact is: \[ V = \frac{m u \cos \alpha (1 + e)}{M + m} \] ### Conclusion The correct option that matches our derived expression is: **Option C:** \( V = \frac{m u \cos \alpha (1 + e)}{M + m} \)

To solve the problem step by step, we will analyze the situation using the principles of conservation of momentum and the coefficient of restitution. ### Step 1: Understand the Setup We have a sphere of mass \( m \) falling towards a smooth hemisphere of mass \( M \). The sphere is moving with a velocity \( u \) just before impact, and the line joining the centers of the sphere and hemisphere makes an angle \( \alpha \) with the vertical. ### Step 2: Resolve the Velocity of the Sphere Before the impact, we can resolve the velocity \( u \) of the sphere into two components: - The vertical component: \( u \cos \alpha \) ...
Promotional Banner

Topper's Solved these Questions

  • ENERGY & MOMENTUM

    VMC MODULES ENGLISH|Exercise LEVEL - 2 PARAGRAPH QUESTIONS|2 Videos
  • ENERGY & MOMENTUM

    VMC MODULES ENGLISH|Exercise JEE MAIN (ARCHIVE)|53 Videos
  • ENERGY & MOMENTUM

    VMC MODULES ENGLISH|Exercise Level - 1 PARAGRAPH QUESTIONS|3 Videos
  • ELECTROSTATICS

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|89 Videos
  • GASEOUS STATE & THERMODYNAMICS

    VMC MODULES ENGLISH|Exercise JEE ADVANCED (ARCHIVE )|111 Videos
VMC MODULES ENGLISH-ENERGY & MOMENTUM-LEVEL - 2
  1. A highly elastic ball moving at a speed of 3 m//s approaches a wall mo...

    Text Solution

    |

  2. Two identical balls are dropped from the same height onto a hard surfa...

    Text Solution

    |

  3. A ball impinges directly on another ball at rest. The first ball is br...

    Text Solution

    |

  4. A mass m moves with a velocity v and collides inelastically with anoth...

    Text Solution

    |

  5. Statement I: In an elastic collision between two bodies, the relative ...

    Text Solution

    |

  6. A sphere of mass m falls on a smooth hemisphere of mass M resting with...

    Text Solution

    |

  7. A ball falls on an inclined plane as shown. The ball is dropped from h...

    Text Solution

    |

  8. An elastic collision takes place between two smooth, rubber balls of s...

    Text Solution

    |

  9. A ball of mass 1 kg moving with velocity vecv = 3hati + 4 hatj collide...

    Text Solution

    |

  10. A ball is projected at t = 0 with velocity v0 at angle theta with ...

    Text Solution

    |

  11. A uniform solid right circular cone has its base cut out in conical sh...

    Text Solution

    |

  12. A square plate of edge d and a circular disc of diameter d are placed ...

    Text Solution

    |

  13. At the same instant that a 0.50-kg ball is dropped from a high buildin...

    Text Solution

    |

  14. From a uniform disc of radius R, an equilateral triangle of side sqrt(...

    Text Solution

    |

  15. In a vertical smooth hollow thin tube, a block of same mass as that of...

    Text Solution

    |

  16. In which of the following cases the centre of mass of a rod is certain...

    Text Solution

    |

  17. A cannon shell is fired to hit a target at a horizontal distance R. Ho...

    Text Solution

    |

  18. A nonzero external force on a system of particles. The velocity and th...

    Text Solution

    |

  19. In the system shown in figure, mass m is released from rest from posit...

    Text Solution

    |

  20. A sphere A is thrown up with velocity 20m/sec from the ground. Sphere ...

    Text Solution

    |