Consider two blocks A and B of masses 
and 
that are placed on a frictionless table. The block A moves with a constant speed 
towards the block B kept at rest. A spring with spring constant 
is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)
Text Solution
Verified by ExpertsB
To solve this problem, we start by using the conservation of linear momentum.
Initially, block A of mass 
is moving with velocity 
, while block B of mass 
is at rest. The blocks move together after the collision. The final velocity of the system is 
. Applying the conservation of linear momentum, we have:
Substituting the known values:
Solving for 
:
Next, we use the conservation of energy to find the compression in the spring. The initial kinetic energy of block A is given by:
The final kinetic energy of both blocks moving
together at 
is:
The difference in kinetic energy is the energy stored in the compressed spring:
Substituting the given spring constant 
into the energy equation:
Solving for 
:
Finally, solving for 
:
Therefore, the compression in the spring is 0.1 m.
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