Question

Constants Part A problem into Find the maximum distance that the block will compress ↑ after the colision- spring after the and apply the appropriate conservation law to each part) (Figure 1),) S 3.00 kg stone traveling whereupon the stone rebounds at 2.00 m/s horizontally to the at 8.00 m/s to the right, Enter your answer using three significant figures X Incorrect; Try Again: 3 attempts remaining Figure 00kg 6.00 ms

Answer 1

Conserving momentum:

m1u1+m2u2 = m1v1 + m2v2

15(0) + 3*8 = 15v + 3*(-2)

v = 2 m/s

Now the initial velocity of the 15 kg block is 2 m/s

KE = (1/2)mv^2 = 1/2 * 15*2*2 = 30 J

KE is converted into spring energy.

Spring energy for 'x' compression = (1/2)kx^2

(1/2)*325*x^2 = 30

x = 0.430 m

Kindly rate :)