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.

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

Answers ( )

  1. mocmien2
    0
    2021-08-21T06:50:27+00:00
    Reply

    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]

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )