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    Bullet collides with mass connected to a spring

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    Two identical wood blocks, A and B, both of mass M, are fastened to the ends of a spring. The spring constant is k. Both blocks are placed on a frictionless horizontal table. Initially, both blocks are at rest and the spring has its unstretched length.

    A bullet of mass m (m<<M) is now fired into A with velocity v, and is absorbed there. This causes A to begin moving in the direction of the spring, toward B. It is assumed that the time it takes the bullet to come to rest is so short that A has not moved appreciably before the bullet has come to rest relative to A. After the bullet is stopped, we ignore all frictional effects. Furthermore, we neglect the mass of the spring. Find the following quantities:

    1) The velocity u of A just after the bullet is stopped relative to A
    2) The total momentum of P of the system following the impact
    3) The velocity v_cm of the center of mass of the system after the impact
    4) The translational kinetic energy E of the system, after the impact (the potential energy die to gravity is constant and may be set equal to zero)
    5) The maximum compression *delta* of the spring
    6) The period T of the oscillations of the spring

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    https://brainmass.com/physics/conservation-of-energy/bullet-collides-with-mass-connected-spring-358142

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    The total momentum the system comprising of the bullet and the two masses A and B, connected by the spring, is conserved. Before the bullet hits the mass, the total momentum is:

    P1 = m v (1)

    After the bullet hits A, and A just starts moving with velocity u when the bullet is stopped, the total momentum is:

    P2 = M u (2)

    Conservation of momentum implies that:

    P1 = P2 ------> (insert (1) and (2) in here)

    m v = M u ----------->

    u = m/M v

    The total momentum of m v divided by the total mass of the system of 2 M gives the center of mass velocity of:

    v_cm = m v/(2M) = 1/2 u

    That this is 1/2 u should be evident from the fact that immediately after impact, A moves with velocity u while B does not move. The center of mass position is the weighted average by mass of the two masses, the center of mass velocity is the time ...

    Solution Summary

    We explain this classical mechanics problem from first principles in different ways.

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