A block of mass 'm' is kept on the wedge PQR of same mass 'm'. All the surfaces are smooth if the system is
released from rest, then choose incorrect option (theta= pi/6, PQ=4 meter, g=10 m/s2).
A
Relative acceleration of block with respect to wedge is 8 m/s2
B
Acceleration of wedge with respect to ground is 2/3 m/s2
С
Time taken by block to reach R from Pis 2 sec
D
Acceleration of block with respect to ground is 28 m/s.
Answers
Answer:
The upper one is the correct and can understand easily
Answer:
Block A of mass m is placed over a wedge of same mass m. Both the block and wedge are placed on a fixed inclined plane. Assuming all surfaces to be smooth, the displacement of the block A in ground frame in 1s is gsin2θx+sin2θ then the value of x is:
check-circle
Text Solution
Answer :
1
Solution :
Let acceleration of wedge in ground frame is a down the plane. The acceleration of block A will be asinθ vertically downward
a→A/g=a→A/B+a→B/g....(i)
[aA/g]x=[aA/B]x+[aB/g]x......(ii)
From F.B.D. of A it is clear that Block A cannot accelerate horizontally i.e., in x-direction because there is no force in x-direction. Block A can accelerate in y-direction only. [aA/g]x=0 Therefore [aA/g]x=−[aB/g]x That means for an observation on wedge block moves only x>0
For block A: mg−N=m(asinθ) ....(iii)
For block B: (N+mg)sinθ=ma ....(iv)
On solving Eqns (iii) and (iv) we get
a=[2gsinθ1+sin2θ]
The acceleration of block A,
aA=asinθ=[2gsinθ1+sinθ]sinθ=[2gsin2θ1+sin2θ]
Displacement of block A in 1 s is
s=0+12aAt2
=12×[2gsin2θ1+sin2θ]×(1)2[gsin2θ1+sin2θ]
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