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 9.00 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 0.150 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 = 5.40 m. What is the value of d?

The figure below shows block 1 of mass 0.200 kg sliding to the right over a frictionless elevated surface at a speed of 9.00 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 0.150 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 5.40 m. What is the value of d? 1 2 k

Explanation / Answer

M2 = kT²/(4²) = 1208.5*(.15)²/(4²) = 0.689 kg
Velocity of center of mass:
Vcm = P/M = 0.200*9.0/(.200+.689) = 2.025 m/s

Velocity of M1 wrt the CM: u1 = 9.0 - Vcm = 6.975 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 = 4.95 m/s

Time to fall 5.4 m: t = [2y/g] = 1.05 sec

d = t*V'1 = 5.2 m

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