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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expression for the
Question
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expression for the minimum value of vm that will allow the second block to circle the loop-the-loop without falling off if the collision is (a) perfectly inelastic or (b) perfectly elastic.
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Physics
1 year
2021-08-21T06:49:13+00:00
2021-08-21T06:49:13+00:00 1 Answers
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Answers ( )
Answer:
Part a) When collision is perfectly inelastic
[tex]v_m = \frac{m + M}{m} \sqrt{5Rg}[/tex]
Part b) When collision is perfectly elastic
[tex]v_m = \frac{m + M}{2m}\sqrt{5Rg}[/tex]
Explanation:
Part a)
As we know that collision is perfectly inelastic
so here we will have
[tex]mv_m = (m + M)v[/tex]
so we have
[tex]v = \frac{mv_m}{m + M}[/tex]
now we know that in order to complete the circle we will have
[tex]v = \sqrt{5Rg}[/tex]
[tex]\frac{mv_m}{m + M} = \sqrt{5Rg}[/tex]
now we have
[tex]v_m = \frac{m + M}{m} \sqrt{5Rg}[/tex]
Part b)
Now we know that collision is perfectly elastic
so we will have
[tex]v = \frac{2mv_m}{m + M}[/tex]
now we have
[tex]\sqrt{5Rg} = \frac{2mv_m}{m + M}[/tex]
[tex]v_m = \frac{m + M}{2m}\sqrt{5Rg}[/tex]