The figure below shows block 1 of mass 0.200 kg sliding to the right over a fric

Question

The figure below shows block 1 of mass 0.200 kg sliding to the right over a frictionless elevated surface at a speed of 6.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.110 s, and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h = 3.50 m. What is the value of d?

Explanation / Answer

We know that, w = sqrt(k/m) and w = 2pi/T

m = k/w^2

For block 2,

M2 = kT^2/(4pi^2)

= 1208.5*0.11^2/(4*pi^2) = 0.37 kg

Velocity of center of mass:

Vcm = P/M = 0.2*6/(0.2+0.37) = 2.10 m/s

Velocity of M1 wrt the CM: u1 = 6 - Vcm = 6 - 2.10 = 3.9 m/s

In elastic collisions, the 2 masses retreat from the CM at the same speed they approached it, and Vcm is constant before & after:

V'1 = u1 - Vcm = 3.9 - 2.1 = 1.8 m/s

Time to fall 3.50 m: t = sqrt(2y/g) = sqrt(2*3.5/9.8) = 0.845 s

d = t*V'1 = 0.845*1.8 = 1.52 m

Answer

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