Question

In the figure, block 2 of mass 2.90 kg oscillates on the end of a spring in SHM with a period of 26.00 ms. The position of the block is given by x = (0.700 cm) cos(?t + ?/2). Block 1 of mass 5.80 kg slides toward block 2 with a velocity of magnitude 8.70 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 6.50 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

Answer 1

Block 1 :

Mass

Velocity

Block 2 :

Mass

Time period

Position

Time instant of collision

Position of block 2 at this time,

Velocity of the block 2 at this time,

Since the collision is completely inelastic, the blocks move with a common velocity after collision.

Spring constant of the spring

At the instant of collision, the mass is at rest at the extreme position, with a displacement .

Now at the same position, the combined block of mass is moving with a speed, .

Conserving the energy of combined mass after colision,