Two blocks can collide in a one-dimensional collision. The block on the left has

Question

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.50 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.60 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use + to mean to the right, - to mean to the left.)

(a) What is the velocity of the first block after the collision?_____ m/s
(b) What is the velocity of the second block after the collision? _____m/s
Remember that the blocks cannot pass through each other!

Explanation / Answer

for elastic collisions

according to conservation of linear momentum

m1*u1 + m2*u2 = m1*v1+m2*v2

m1*(u1-v1) = m2*(v2-u2)........(1)

according to conservation of energy

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

m1*(u1^2 - v1^2) = m2*(v2^2-u2^2).....(2)

from 1 &2

u1 + v1 = u2+v2

u1 - u2 = v2 - v1

v2 = u1 - u2 + v1........(3)

3 in 1

m1*(u1-v1) = m2*(u1 - u2 + v1 - u2)

v1 = u1*(m1-m2)/(m1+m2) + 2*m2*u2/(m1+m2)

v2 = u2*(m2-m1)/(m1+m2) + 2*m1*u1/(m1+m2)

m1 = 0.5 kg.....m2 = 0.6 kg

u1 = +2.4 m/s ...........u2 = 0 m/s

a) v1 = (2.4*(0.5-0.6))/(0.5+0.6) = - 0.2182 m/s < answer (towards left)

b) v2 = (2*0.5*2.4)/(0.5+0.6) = + 2.182m/s < ------answer(towards right)

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