Question

The figure shows a block of mass m placed on a smooth wedge of mass M. The value of M' is just sufficient, so that the block of mass m moves vertically downwards with acceleration 10 m/$s^2$_. Chose the correct option:

• A. This is possible only for
• B. This is possible only for
• C. If This is possible then the value of M' is
• D. If This is possible then the value of tension in the string is

Answer 1

Answer: (B), (C), (D)

The correct options are
B

This is possible only for

C

If This is possible then the value of M' is

D

If This is possible then the value of tension in the string is

Let's draw the free body drawing of block m

Hmmm...........

The direction of net acceleration of the block should be the direction of the net force but that doesn't seem to be the case.

So N has to be O

and mg = m a

a = g = 10 m/$s^2$

correct ! ! !

so there is no Normal between block and wedge (no contact). Now the question very carefully says that the block has just reached this situation. So the contact is just lost or on the verge of getting lost.

We can still apply our wedge constraint and see what's the relation between the acceleration.

$M^{1}g-T=M^{1}a$ -----------------(i)

T = Ma ----------------(ii)

Solving (i) & (ii) $a=\frac{M^{1}g}{(M^1+M)}$ ...(iii)

Constraint relation between acceleration of block of mass m and wedge of Mass M: acceleration in perpendicular direction to surface of wedge will be equal for both wedge and block.

asin$\theta$ = g cos$\theta$

a = g cot $\theta$

substituting in eg (iii)

$\frac{M^{1}g}{(M^1+M)}=gcot\theta$

$tan\theta =(1+\frac{M}{M^1})$---------(iv)

so $tan\theta >1$

$\Rightarrow \theta >{45}^{\circ }$

From eq...(iv)

$M^1=\frac{Mcot\theta }{1-cot\theta }$

Substituting value of a in eq --------(ii)

$T=\frac{Mg}{tan\theta }$

so b, c, d