A 14 gram clay mass moving at 50 m/s collides into, and then sticks to, an 86 gr

Question

A 14 gram clay mass moving at 50 m/s collides into, and then sticks to, an 86 gram clay mass at rest. There are no external forces acting on these masses during the collision.

A quantity is said to be conserved during a collision if that quantity after the collision is equal to its value before the collision.

a.What is the speed in m/s of the resulting mass post collision?

b. What is the ratio of the post-collision kinetic energy of the system to the pre-collision kinetic energy of the system (K_post/K_pre) ?

Explanation / Answer

here,

mass of clay A , ma = 0.014 g

mass of clay B , mb = 0.086 kg

ua = 50 m/s

a.

let the speed of the resulting mass be v

using conservation of momentum

ma * ua = ( ma + mb) * v

0.014 * 50 = ( 0.014 + 0.086) * v

v = 7 m/s

the speed of resulting mass is 7 m/s

b.

the ratio of post-collision kinetic energy of the system to the pre-collision kinetic energy of the system , Kpost / Kpre = ( 0.5 * ma*ua^2) /( 0.5 * ( ma + mb) * v^2)

Kpost / Kpre = 0.014 * 50^2 /( 0.1 * 7^2)

Kpost / Kpre = 7.14

the ratio of post-collision kinetic energy of the system to the pre-collision kinetic energy of the system is 7.14 :1

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