Question
A sphere of mass 500g moving with a velocity of 200 cm s-1collides centrally with another sphere of mass 100 g moving with a velocity of 100 cm s-1towards it. After the collision the two spheres stick together. Find the final velocities of the two spheres and the loss in kinetic energy of the system.
Plz answer fast
Plz answer fast
Answers
Damon
use meters and kilograms or I get confused
original momentum = 0.5 * 0.2 - 0.1 * 0.1= 0.09 kg m/s
final momentum = 0.6*v
final momentum = original so
0.6 v = 0.09
v = 0.15 m/s
original ke = (1/2)[0.5*(0.2)^2 + 0.1(0.1)^2] = .5[.02+.001]
= 0.0105 Joules
final ke =.5[0.6 *(0.15)^2] = 0.00675 Joules
subtract them
original momentum = 0.5 * 0.2 - 0.1 * 0.1= 0.09 kg m/s
final momentum = 0.6*v
final momentum = original so
0.6 v = 0.09
v = 0.15 m/s
original ke = (1/2)[0.5*(0.2)^2 + 0.1(0.1)^2] = .5[.02+.001]
= 0.0105 Joules
final ke =.5[0.6 *(0.15)^2] = 0.00675 Joules
subtract them
henry2,
Given: M1 = 0.5 kg, V1 = 2 m/s.
M2 = 0.1kg, V2 = 1m/s.
V3 = Velocity 0f M1 and M2 after collision.
Momentum bef0re = Momentum after.
M1*V1+M2*V2 = M1*V3+M2*V3
0.5*2+0.1*(-1) = 0.5V3+0.1V3
0.6V3 = 0.9
V3 = 1.5 m/s.
KEb = 0.5M1*V1^2+0.5M2*V2^2 = 0.25*2^2+0.05*(-1)^2 = 1.05 Joules.
KEa = 0.5M1*V3^2+0.5M2*V3^2 = 0.25*1.5^2+0.05*1.5^2 = 0.675 Joules.
KE lost = KEb-KEa =
M2 = 0.1kg, V2 = 1m/s.
V3 = Velocity 0f M1 and M2 after collision.
Momentum bef0re = Momentum after.
M1*V1+M2*V2 = M1*V3+M2*V3
0.5*2+0.1*(-1) = 0.5V3+0.1V3
0.6V3 = 0.9
V3 = 1.5 m/s.
KEb = 0.5M1*V1^2+0.5M2*V2^2 = 0.25*2^2+0.05*(-1)^2 = 1.05 Joules.
KEa = 0.5M1*V3^2+0.5M2*V3^2 = 0.25*1.5^2+0.05*1.5^2 = 0.675 Joules.
KE lost = KEb-KEa =