Question

1.An elastic wire extend by 1.0cm when a load of 20g hangs from it.what additional load will be required to cause a further extension of 2.0cm?
2.A ball A of mass 0.4kg moving with velocity 5ms-1 collides head on with a ball B of mass 0.2kg moving with velocity 2ms-1, after the collision,it move with the velocity 3ms-1 and B with velocity 6ms-1
(a) Show that collision obeys the law of conversation of momentum.
(b) Is the collision elastic or inelastic.?

Answer 1

1. 20g/1cm = (20g+x)/2cm.
20/1 = (20+x)/2.
X = ?

2. Given:
M1 = 0.4 kg, V1 = 5 m/s.
M2 = 0.2 kg, V2 = -2 m/s.
V3 = 3m/s = velocity of M1 after collision.
V4 = 6m/s = velocity of M2 after collision.

a. Momentum before collision:
M1*V1 + M2*V2 = 0.4*5 + 0.2*(-2) = 1.6 kg-m/s.

Momentum after collision:
M1*V3 + M2*V4 = 0.4*3 + 0.2*6 = 2.4kg-m/s.
The Momentum before and after collision should be equal. but they are not
unless we assume the 2m/s is positive. But it should be negative, because
the balls are moving in the opposite direction. Please check the given data
for errors.

Answer 2

I recalculated the velocities after the collision :
V3 = 1/3 m/s = velocity of M1 after the collision.
V4 = 22/3 = 7 1/3 m/s = velocity of M2 after the collision.

a. Momentum before the collision:
M1*V1 + M2*V2 = 0.4*5 + 0.2*(-2) = 1.6 kg-m/s.

Momentum after the collision:
M1*V3 + M2*V4 = 0.4*1/3 + 0.2*(22/3) = 4.8/3 = 1.6 kg-m/s.

b. During an elastic collision, momentum AND KE are conserved.
KE before the collision:
0.5M1*V1^2 + 0.5M2*V2^2 = 0.2*5^2 + 0.1*(-2)^2 = 5 + 0.4 = 5.4 Joules.

KE after the collision:
0.5M1*V3^2 + 0.5M2*V4^2 = 0.2*(1/3)^2 + 0.1*(22/3)^2 = 5.4 Joules.

Answer 3

Yes

Answer 4

I do not under stand the question