Question

A block B of mass 0.6kg slides down the smooth face PR of wedge A of mass 1.7 kg which can move freely on a smooth horizontal surface. The inclination of the face PR to the horizontal is 45 ^•.Then:

A)The acceleration of A is 3g/20

B)the vertical component of the acceleration of B is 23g/40

C)the horizontal component of the acceleration of B is 17g/40

D)None of these

Answer 1

Answer:

Explanation:

Mass of the wedge M = 1.7 kg  (Given)

Mass of the block m = 0.6 kg  (Given)

On A, a force N cos 45° acts towards the left,

where N is the contact force between A and B acting as perpendicular to the face of the incline. Thus, if a0 be A's acceleration towards the left -

N sin 45° = M a0 ... (1)

Since, B slides down the incline. Thus, -  

mg sin45° + ma0 cos45° = ma' ... (2)

When normal to the face: mg cos45° = N + ma0 sin45° ...(3)

Solving for a0, a' -

a0 = mm+2 M g = .64 g

= 3/20 g

And a' = √2 (m+M)(m+2M) g

= 23√240 g

Thus, the vertical component of B's acceleration is a' sin 45° = 2340 g

The horizontal acceleration of B -

a' cos 45° - a0

= 2340 g - 320 g

= 1720 g

Thus, all three statement a), b), c) are correct.

Answer 2

Answer:

Explanation:

Mass of the wedge M = 1.7 kg (Given)

Mass of the block m = 0.6 kg (Given)

On A, a force N cos 45° acts towards the left,

where N is the contact force between A and B acting as perpendicular to the face of the incline. Thus, if a0 be A's acceleration towards the left -

N sin 45° = M a0 ... (1)

Since, B slides down the incline. Thus, -

mg sin45° + ma0 cos45° = ma' ... (2)

When normal to the face: mg cos45° = N + ma0 sin45° ...(3)

Solving for a0, a' -

a0 = mm+2 M g = .64 g

= 3/20 g

And a' = √2 (m+M)(m+2M) g

= 23√240 g

Thus, the vertical component of B's acceleration is a' sin 45° = 2340 g

The horizontal acceleration of B -

a' cos 45° - a0

= 2340 g - 320 g

= 1720 g

Thus, all three statement a), b), c) are correct.