An ideal spring is permanently connected between two blocks of masses and . The blocks-spring system can move over a smooth horizontal table along a straight line along the length of the spring as shown in Fig. The blocks are brought nearer to compress the spring and then released. In the subsequent motion

a) |
Initially they move in opposite directions with velocities inversely proportional to their masses |

b) |
The ratio of their velocities remains constant |

c) |
Linear momentum and energy of the system remain conserved |

d) |
The two blocks will oscillate about their centre of mass, which remains stationary |

An ideal spring is permanently connected between two blocks of masses and . The blocks-spring system can move over a smooth horizontal table along a straight line along the length of the spring as shown in Fig. The blocks are brought nearer to compress the spring and then released. In the subsequent motion

a) |
Initially they move in opposite directions with velocities inversely proportional to their masses |

b) |
The ratio of their velocities remains constant |

c) |
Linear momentum and energy of the system remain conserved |

d) |
The two blocks will oscillate about their centre of mass, which remains stationary |

1 Answer

127 votes

Since the system of two blocks and a spring is placed on a smooth horizontal floor, therefore no external horizontal force acts on the system. Since the system is initially at rest, therefore centre of mass of the system will remain stationary and the block will oscillate with the same frequency. It means option (d) is correct. Since the system, is initially at rest and centre of mass stationary, therefore centre of mass cannot move or the momentum of the system remains equal to zero. It means, momenta of two blocks are numerically equal but they are in opposite directions. Therefore, ratio of magnitude of velocity of the two blocks will be in inverse ratio of their masses. Hence, option (a) is correct.Since ratio of mass is constant, therefore the ratio of their velocity will remain constant. It means that option (b) is also correct Since the system is on smooth horizontal floor, therefore no energy loss takes place. Hence, linear momentum and energy of the system remain conserved. Therefore, option (c) is also correct |

127 votes

127