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Question

In the figure shown , the acceleration of A is aA=(15^i+15^j) m/s2 and block A remains in contact with wedge B. The acceleration of B (in m/s2) will be.


A
6^i
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B
15^i
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C
10^i
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D
5^i
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Solution

The correct option is D 5^i
Let us suppose, the wedge move towards +ve x-axis


From wedge constraints the components of acceleration of A and B along the normal direction must be equal.

(aA)=(aB)

(aA)xcos53(aA)ycos37=aBcos53...(1)

Given,
aA=(15^i+15^j) m/s2

So, (aA)x=15 m/s2, (aA)y=15 m/s2

Substituting the values in equation (1),

(15×35)(15×45)=aB(35)

912=3aB5

aB=5

Therefore wedge will move along -ve x-direction.

aB=5^i m/s2

Hence, option (d) is the correct answer.
Why this question?
Tip:-When two rigid bodies are in contact, the component of acceleration perpendicular to contact surface must be same, because there is no relative motion between bodies along that direction.

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