Question

A metal sphere with a mass of 80.0g rolls along frictionless surface at 20.0m/s and strikes a stationary sphere having a mass of 200.0g. The First sphere stops completely. At what speed does the second sphere move away from the point of impact

...?

Answer 1

8.0 m/s

Step-by-step explanation:

By the law of conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision:

`p_i=p_f`

The total momentum before the collision is given only by the momentum of the first sphere, since the second sphere is stationary (so its speed is zero and its momentum is zero as well):

`p_i = m_1 u_1 =(80.0 g)(20.0 m/s)=1600 g m/s`

The total momentum after the collision is given only by the momentum of the second sphere, since the first sphere completely stops, so:

`p_f = m_2 v_2`

Using conservation of momentum, we find

`p_i=m_2 v_2\\1600 g m/s = (200.0 g) v_2 \\v_2 = (1600 g m/s)/(200.0 g)=8.0 m/s`

Answer 2

To know what speed does the second sphere move away from the point of impact, use preservation of momentum. Total momentum will stay the same before and after the impact.

m1v1=m2v2. The first sphere will stop completely after the impact.
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