Question

16. A small block A of mass mo is given a velocity v in
horizontal direction as shown. The bigger block B is
initially at rest. The curved surface of the block B
becomes vertical at the top point if M = 4m, what
should be the minimum value of v so that the block
A climbs upto the top on B? All surfaces are
smooth (g = 10 m/s2)

(1) 20 m/s
(3) 9 m/s
(2) 10 m/s
(4) 2 m/s​

Answer 1

Explanation:

THE ANSWER OF THE ABOVE QUESTION.

Answer 2

Answer:

The correct answer for this question is Option(2).

Explanation:

Given: A small block A of mass is given a velocity (v) in a horizontal direction. The bigger block B is initially at rest. The curved surface of the block B becomes vertical at the top point if M = 4m,

To find: what should be the minimum value of v so that the block A climbs upto the top on B

Applying linear momentum,

$M_{o}V$ + 0 = $(M_{o} + 4M_{o} ) V'$

V' = $\frac{M_{o}V }{5M_{o} }$ = $\frac{V}{S}$

Now following the L.O.C.O energy on the equation:

$\frac{1}{2}M_{o} V^{2}$ = $M_{o}g * 4$ + $\frac{1}{2}$ *$\frac{V^{2} }{25}$ $(4M_{o} + M_{o})$

$\frac{V^{2} }{2}$ = 40 + $\frac{V^{2} }{2} * \frac{5}{25}$

$\frac{V^{2} }{2} * \frac{1}{5}$

$V^{2} = 80 + \frac{V^{2} }{5}$

$\frac{4V^{2} }{5}$ = 80

$V^{2} = 100$

V= 10 m/s.

Therefore the minimum value of v so that the block A climbs upto the top on block  B is 10m/s.

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