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Question

A block released from the state of rest from a wedge as shown. The block and the wedge are of the same mass. If all surfaces are smooth, the total horizontal displacement of the block with respect to earth before it falls on the earth is given by xR. Then x=

Take √2=1.41

Take √2=1.41

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Solution

Since, x-coordinate of center of mass remains same.

xi=m(a)+m(a1)2m=a+a12

Let x is the displacement of wedge when block reaches the bottom.

xf=m(a−x)+m(a1+R−x)2m =a+a1+R−2x2

∴xi=xf i.e. x=R2=0.5 R

After that particle will follow the parabolic path

R=ut+12gt2

where,u=0 m/s

so,

R=12gt2

∴t=√2Rg

Let x′ is the displacement of the block when it reaches the surface and Vm is the initial velocity of the block.

x′=vt=√2Rg.Vm

From momentum conservation

mVm=mVM (where Vm and VM are velocity's of the block and the wedge when the block just leaves the wedge)

Hence Vm=VM

Using conservation of energy:

mgR=12mV2m+12mV2m

mgR=mV2m

Vm=√gR

Using eq. (1)

x′=√2Rg×√gR

∴x′=√2R

∴Total displacement=x+x′ =0.5R+√2 R

=(0.5+1.41)R =1.91R

xi=m(a)+m(a1)2m=a+a12

Let x is the displacement of wedge when block reaches the bottom.

xf=m(a−x)+m(a1+R−x)2m =a+a1+R−2x2

∴xi=xf i.e. x=R2=0.5 R

After that particle will follow the parabolic path

R=ut+12gt2

where,u=0 m/s

so,

R=12gt2

∴t=√2Rg

Let x′ is the displacement of the block when it reaches the surface and Vm is the initial velocity of the block.

x′=vt=√2Rg.Vm

From momentum conservation

mVm=mVM (where Vm and VM are velocity's of the block and the wedge when the block just leaves the wedge)

Hence Vm=VM

Using conservation of energy:

mgR=12mV2m+12mV2m

mgR=mV2m

Vm=√gR

Using eq. (1)

x′=√2Rg×√gR

∴x′=√2R

∴Total displacement=x+x′ =0.5R+√2 R

=(0.5+1.41)R =1.91R

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