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

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 Vodka 1 year 2021-08-21T06:49:13+00:00 2021-08-21T06:49:13+00:00 1 Answers 369 views 0

Answers ( )

mocmien2 · Answer 1 · 2021-08-21T06:50:27+00:00

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]