Physics; Momentum; Not Homework?

Question

particles A and B, of masses 0.15kg and 0.2kg respectively, are free to move on a horizontal surface. air resistance may be ignored. at a particular instant A is moving with speed 2ms-1 towards B, which is stationary at a point 4m from A. particle A collides directly with particle B.

it is given instead that the coefficient of friction between A and the surface is 0.05. A is again brought to rest by the collision. Find the speed of B immediately after the collision?

Answer 1

V = 2 m/s
Ma = 0.15 kg
Mb = 0.20 kg
d = 4 m
g = 9.8 m/s²
k = 0.05

Initial energy of particle A: K0 = ½ Ma V²
Loss of energy due to friciton: ΔE = k Ma g d
Kinetic energy before impact: Ka = Ma (½ V² - kgd)
Momentum before impact: P1 = √(2 Ma Ka) = Ma √(V² - 2kgd)
Momentum after impact: P2 = Mb Vb = P1 = Ma √(V² - 2kgd)
Speed of B immediately after the collision:

Vb = Ma/Mb √(V² - 2kgd) = 20 cm/s

The final result is very inaccurate, because the first
particle A almost comes to a stop before the collision.

Answer 2

First you need to figure out what the speed of particle A was just before the collision. You know it is being decellerated by a constant force F=mu*Fn=mu*Fg=mu*m*g where mu is the coefficient of kinetic friction (funny looking u). solving for a you get a=mu*g. Now just apply your knowledge of kinematics to find the final speed. The last step is just to say that the momentum before the collision is equal to the momentum afterwards m_A*v_A=m_B*v_B. Good luck!

Answer 3

kinetic energy