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Question

Blocks A and B are at rest with the spring at its natural length and block C collides elastically with A. Then,

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Solution

The correct option is **D** The maximum compression of the spring is 1 m.

From the data given in the question,

Block C moving with velocity 20 m/s upon colliding with A elastically, will come to rest instantly and block A will start moving with 20 m/s.

At maximum compression, blocks A and B will move with a common velocity equal to the velocity of their centre of mass as shown in the figure below.

Let us suppose, the velocity of COM be v0 and maximum compression of the spring be x.

According to question,

From figure II:

vCOM=v0=1×20+1×01+1=10 m/s−−−(1)

[∵vCOM=m1v1+m2v2m1+m2]

We know that in collision, only internal forces operate and hence ∑Fint=0. Therefore, there is no net force on system A.B.

Thus, VCOM will not change.

On applying law of conservation of mechanical energy between blocks A and B as shown in figure II and III,

M.Ei=M.Ef

⇒M.EII=M.EIII

Initial velocity of Block A (u2)=20 m/s

Initial velocity of Block B (u3)=0 m/s

Final velocity of Block A (v2)=10 m/s

Final velocity of Block B (v3)=10 m/s

12mu22+0=12mv22+12mv23+12kx20

⇒12×1×202+0=12×1×102+12×1×102+12×200×x20

⇒x0=1 m is the maximum compression in the spring.

Kinetic energy of A - B system at maximum compression.

=12×1×102+12×1×102

K.E=100 J

Options B and D are correct.

From the data given in the question,

Block C moving with velocity 20 m/s upon colliding with A elastically, will come to rest instantly and block A will start moving with 20 m/s.

At maximum compression, blocks A and B will move with a common velocity equal to the velocity of their centre of mass as shown in the figure below.

Let us suppose, the velocity of COM be v0 and maximum compression of the spring be x.

According to question,

From figure II:

vCOM=v0=1×20+1×01+1=10 m/s−−−(1)

[∵vCOM=m1v1+m2v2m1+m2]

We know that in collision, only internal forces operate and hence ∑Fint=0. Therefore, there is no net force on system A.B.

Thus, VCOM will not change.

On applying law of conservation of mechanical energy between blocks A and B as shown in figure II and III,

M.Ei=M.Ef

⇒M.EII=M.EIII

Initial velocity of Block A (u2)=20 m/s

Initial velocity of Block B (u3)=0 m/s

Final velocity of Block A (v2)=10 m/s

Final velocity of Block B (v3)=10 m/s

12mu22+0=12mv22+12mv23+12kx20

⇒12×1×202+0=12×1×102+12×1×102+12×200×x20

⇒x0=1 m is the maximum compression in the spring.

Kinetic energy of A - B system at maximum compression.

=12×1×102+12×1×102

K.E=100 J

Options B and D are correct.

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