Three objects are positioned as shown in the diagram below. Object 1 collides wi

Question

Three objects are positioned as shown in the diagram below. Object 1 collides with Object 2 perfectly inelastically, and then Object 2 collides with Object 3 elastically. Object 3 then collides with a spring.

a.What is the velocity of Object 1 just prior to the collision with Object 1?

b.What is the velocity of Object 2 just prior to colliding with Object 3?

c.What is the velocity of Object 2 just after colliding with Object 3?

d.How far will the spring be compressed when it is struck by Object 3, assuming that the spring has a spring constant of 10,000 N/m?

Explanation / Answer

Given,

h = 2 m ; m1 = 1kg ; m2 = 2 kg ; m3 = 3kg

a)The gravitational PE of object will get converted to KE of motion, So

m1 g h = 1/2 m1 v12

v1 = sqrt (2 g h) = sqrt (2 x 9.8 x 2) = 6.26 m/s

Hence, the velocity of Object 1 just prior to the collision with Object 2 is v1 = 6.26 m/s.

b)we have v1 = 6.26 m/s and v2 = 0

As its a perfectly ineastic collsion, the masses will be clubbed into one and move with same speed.

we know that, momentum will be conserved so

Pi = Pf

m1v1 + m2v2 = (m1 + m2) Vf

m1v1 = (m1 + m2)Vf

Vf = m1v1/(m1+m2) = 1 x 6.26 / (1 + 2) = 2.087 m/s

Hence, Vf = 2.087 m/s.

c)Now m1+m2 can is already a single mass moving with velocity Vf = 2.087 m/s

v3 = 0

Again all the masses would combine to become one moving mass, so the velocity of m1, m2 and m3 will be same. Again from momentum conservation

(m1+m2)Vf + m3v3 = (m1 + m2 + m3) Vf'

Vf' = (m1 + m2) Vf / (m1 + m2 + m3)

Vf' = (1 + 2 ) x 2.087 / ( 1 + 2 + 3) = 1.044 m/s

Hence, Vf' = 1.044 m/s

(d)k = 10,000 N/m

From energy conservation, the kinetic energy of the moving masses will get converted to the spring potential energy, so

KE = PE

1/2 m Vf'2 = 1/2 k x2

where, x is the compression of the spring and m = m1 + m2 + m3 = 6 kg

x = Vf' sqrt (m/k) = 1.044 x sqrt (6/100000) = 0.0256 m

Hence, x = 0.0256 m = 2.56 cm

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