Question

In the figure, block 2 of mass 2.60 kg oscillates on the end of a spring in SHM with a period of 24.00 ms. The position of the block is given by x = (1.60 cm) cos(wt + pi/2). Block 1 of mass 5.20 kg slides toward block 2 with a velocity of magnitude 7.20 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 6.00 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

Given that,

m2 = 2.60 ; t = 24ms = 24 x 10-3 s ;

x = (1.60cm) cos( $\omega$ t + pi/2); m1 = 5.20 kg v1 = 7.20 m/s

We know that

$\omega$ = 2 pi/t = 2 x 3.14 / 24 x 10-3 ms = 261.7 rad/sec

k = $\omega$ ² x m2 = (261.7)² x 2.6 = 1.78 x 10⁵ N/m

Intially the block 2 was in rest and 1 was moving with velocit v1, so system's intial momentum is due to block 1 only.Since the collsion is inelastic, the two masses after the collision got sticked together, So using linear momentum conservation, we can write for their combines speed as(let v be the velocity after collision)

m1v1 = (m1+m2)v

v = m1v1/(m1+m2) = 5.2 x 7.2 / (5.2 + 2.6) = 37.44 / 7.8 = 4.8 m/s

There will be the net KE of the system due to the gain in speed and the Net PE of the spring. So

KE = 1/2(m1+m2)v² = 1/2 (5.2 + 2.6) (4.8)² = 89.86 Joules

PE(U) = 1/2 K x² = 1/2 (1.78 x 10⁵ N/m) x (0.016)² =22.78 Joules

Total mechanical energy of the system = KE + PE = 89.86 + 22.78 = 112.64 Joules

Let X be the new amplidure of the system, Then we can write for PE as;

PE = 1/2 k X²

X = sqrt ( 2 x PE / k ) = sqrt ( 2 x 112.64 / 1.78 x 10⁵ N/m ) = 0.0356 m

Hence, The amplitude of SHM after the collision occured = X = 0.0356 meters