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    A softball of mass 0.221 kg that is moving with a speed of 6.5 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward with a speed of 4.3 m/s.

    (a) Calculate the velocity of the target ball after the collision.

    (b) Calculate the mass of the target ball.

    © BrainMass Inc. brainmass.com December 24, 2021, 8:08 pm ad1c9bdddf
    https://brainmass.com/physics/conservation-of-momentum/collision-two-balls-251641

    SOLUTION This solution is FREE courtesy of BrainMass!

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    SOLUTION

    Principles used :

    1. Law of conservation of linear momentum : When two bodies collide, their momentum is conserved. That is, total momentum of the bodies before and after collision is same. This is applicable irrespective of whether the collision is elastic or inelastic.

    2. Conservation of kinetic energy : In an elastic collision, the total kinetic energy of the colliding objects is conserved. That is total KE before and after collision is same. (If the collision is inelastic, some kinetic energy is dissipated in the process of deformation of the colliding bodies. Thus, KE is not conserved)

    (See attached file for diagram)

    Applying conservation of momentum we get :

    m1u1 + m2u2 = m1v1 + m2v2 ............(1) [where u1, u2 are velocities before collision and v1, v2 after collision]

    As the collision is elastic, we can apply the law of conservation of KE :

    ½ m1(u1)2 + ½ m2(u2)2 = ½ m1(v1)2 + ½ m2(v2)2

    m1(u1)2 + m2(u2)2 = m1(v1)2 + m2(v2)2 .............(2)

    Solving (1) and (2) for v1 and v2 we get :

    v1 = (m1 - m2)u1/(m1+m2) + 2m2u2/(m1+m2)

    v2 = (m2 - m1)u2/(m1+m2) + 2m1u1/(m1+m2)

    Putting values as follows :

    m1 = 0.221 kg, u1 = +6.5 m/s, u2 = 0, v1 = -4.3 m/s [velocity towards right is taken as +ve and that towards left is taken as -ve.]

    -4.3 = (0.221 - m2) 6.5/(0.221 + m2) + 2 x m2 x 0/(0.221 + m2)

    Or 1.44 - 6.5m2 = - 0.95 - 4.3m2

    Or m2 = 2.39/2.2 = 1.08 kg

    v2 = (1.08 - 0.221)(0) /(1.08+0.221) + 2 x 0.221 x 6.5/(0.221+1.08)

    Or v2 = 2 x 0.221 x 6.5/(0.221+1.08) = 2.2 m/s

    Mass of the target ball is 1.08 kg and its after collision velocity is 2.2 m/s towards right (+ve).

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 8:08 pm ad1c9bdddf>
    https://brainmass.com/physics/conservation-of-momentum/collision-two-balls-251641

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