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.

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)