Two objects of masses m 1 = 0.40 kg and m 2 = 0.92 kg are placed on a horizontal

Question

Two objects of masses

m1 = 0.40 kg

and

m2 = 0.92 kg

are placed on a horizontal frictionless surface and a compressed spring of force constant

k = 290 N/m

is placed between them as in figure (a). Neglect the mass of the spring. The spring is not attached to either object and is compressed a distance of 9.6 cm. If the objects are released from rest, find the final velocity of each object as shown in figure (b). (Let the positive direction be to the right. Indicate the direction with the sign of your answer.)

Explanation / Answer

let v1 and v2 are the speeds of m1 and m2 after they move they from each other.

Apply conservation of momentum

m1*v1 = m2*v2

0.04*v1 = 0.92*v2 ----(1)

Apply conservation of energy

initial mechanical energy = final mechanical energy

(1/2)*k*x^2 = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

(1/2)*290*0.096^2 = (1/2)*0.4*v1^2 + (1/2)*0.92*v2^2 ---(2)

on solving the above two equations we get

v1 = 2.58 m/s
v2 = 0.112 m/s

so, if m1 is towards right

the answers are
v1 = 2.58 m/s
v2 = -0.112 m/s

if m2 is towards right
the answers are
v1 = -2.58 m/s
v2 = 0.112 m/s

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