COLLISSION SOMEONE PLEASE HELP IN PHYSICS!!?

Question

1) A ball of mass 0.255kg that is moving with a speed of 7.7m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.7m/s . 

A) Calculate the velocity of the target ball after the collision. 

B) Calculate the mass of the target ball. 

2) The jet engine of an airplane takes in 120 kg of air per second, which is burned with 4.6kg of fuel per second. The burned gases leave the plane at a speed of 600m/s (relative to the plane). If the plane is traveling 290m/s (650mi/hr ), determine the following quantities. 

A) The thrust due to accelerated air passing through the engine.

JullyWum · Accepted Answer

(1) Let the mass of target = m2 and it's velocity after impact v2
Applying cons of momentum .. motion in same initial direction as incoming ball taken as +ve 
Total mom before impact = total afterwards
(0.255kg x 7.70m/s) + 0 = (0.255kg x -3.70m/s) + (m2 x v2)
m2.v2 = 2.907 kg.m/s ---- (A)

Applying the relative velocity rule to elastic collision ..
(u1 - u2) = (v2 - v1)
7.70 - 0 = v2 - [-3.7]) .. .. ►v2 = (+) 4.0 m/s

1B) From equ (A) above, mom of target = 2.907 = m2 x v2

m2 = 2.907kg.m/s / 4.0m/s .. .. ► m2 = 0.727 kg

2A) 
Thrust (reaction force) = rate of change of momentum of exhaust
= (m∆v)/t = (m/t)kg/s x ∆v(m/s)

= (120.0 + 4.6)kg/s x (600 - 290)m/s  .. .. ►Thrust = 3.86^4 N

Madison · Answer

The first one: both kinetic energy and momentum are conserved. 
Use conservation of kinetic energy first to solve for the new velocity:
(V1-V2)intial= -(V1-V2)final 

Then, use your new velocity and all your known values in the conservation of momentum problem:
(m1v1 + m2v2) intial = (m1v1 + m2v2) final

Note: your v1(final) is negative (-3.7 m/s) because it bounces backwards. That will affect your results.

COLLISSION SOMEONE PLEASE HELP IN PHYSICS!!?

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