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.

Answer 1

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Answer 2

Following are the calculations to the perfectly inelastic and perfectly elastic.

For point a) perfectly inelastic:

• When demand is perfectly relatively elastic, the overall quantity demanded for a good does not vary in response to changes in price.
• Ultimately, whenever the PED coefficient equals infinity, desire is said to be perfectly elastic.

Since we all know, collisions are totally inelastic:

$m \ v_m=(m+M)v$

$v=\frac{mv_m}{m+m}$

In order to complete the circle:

For point b) perfectly elastic:

• When a new price results in an infinite amount of change in quantity, the term "completely elastic".
• When the price falls, the quantity demanded rises to infinity, therefore, as the price rises, the demand curve declines to zero.
• When the collision is perfectly elastic:

          $v=\frac{2mv_m}{m+M}\\\\\sqrt{5Rg}= \frac{2mv_m}{m+M}\\\\v_m=\frac{m+M}{2m} \sqrt{5Rg}$

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