Acollision in one dimension
a mass m1 = 2 kg moving at v1i = 3 ms−1 collides with another mass m2 = 4 kg moving at
v2i = −2 ms−1. after the collision the mass m1 moves at v1f = −3.66 ms−1. (a) calculate the final velocity of the mass m2.
(b) after the collision the mass m1 slides across a surface with coefficient of friction μ = 0.6. calculate how far it travels before it comes to rest.

a) v2f = 0.83 m/s b) The distance will be 0.3738 m.

Explanation: a) When two bodies collide with each other with definite velocity, the system is a closed system, which means, there are no other forces acting on it, and the linear momentum is constant. So, for a system of two bodies:

Qi = Qf

m1v1i + m2v2i = m1v1f + m2v2f

2.3 + 4.(-2) = 2.(-3.66) + 4v2f

v2f =

v2f = 0.83

The final velocity of mass m2 is v2f = 0.83 m/s.

b) Since m1 slides with coefficient of friction, there is a force of friction acting on it:

= μ·, where is a normal force acting on m1.

As there is no up or down movement, = = m.g = 2*9.8

= 0.6*2*9.8

Only the force of friction is acting on m1, so:

= m.a

0.6*2*9.8 = 2*a

a = 5.88 m/s²

Using

d =

d = 0.3738

The mass m1 traveled 0.3738 m before it came to rest.

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can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s