The figure shows block 1 of mass 0.180 kg sliding to the right over a frictionle

Question

The figure shows block 1 of mass 0.180 kg sliding to the right over a frictionless elevated surface at a speed of 7.75 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1133 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.180 s, and block 1 slides off the opposite and of the elevated surface, landing a distance d from the base of the surface after falling height h = 4.50 m. What is the value of d?

Explanation / Answer

Here , m1 = 0.180 Kg

let the mass of second mass is m2

as period = 2pi *sqrt(m/k)

0.180 = 2*pi * sqrt(m2/1133)

solving for m2

m2 = 0.93 Kg

let the final speed of masses is v1 and v2

Using conservation of mometum

0.180 * v1 + 0.930 * v2 = 0.180 * 7.75 ----(1)

for elastic collision ,

coefficient of restitution

v2 - v1 = 7.75 ---(2)

solving 1 and 2

v1 = -5.24 m/s

v2 = 2.51 m/s

NOw, time taken for block 1 to fall ,

tf = sqrt(2h/g)

tf = sqrt(2 * 4.5/9.8)

tf = 0.958 s

distance , d = tf * v1

d = 0.958 * 5.24 m

d = 5.02 m

the distance d is 5.02 m

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