Question

A collision 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.

Answer 1

Step-by-step explanation:

It’s easy. The answer is k=1{%5) -3q* -v1f

Answer 2

Answer: 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 =
`(2.66)/(2)`

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:

`F_(f)` = μ·
`F_(N)`, where
`F_(N)` is a normal force acting on m1.

As there is no up or down movement,
`F_(N)` =
`F_(g)` = m.g = 2*9.8

`F_(f)` = 0.6*2*9.8

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

`F_(f)` = m.a

0.6*2*9.8 = 2*a

a = 5.88 m/s²

Using
`(v_(f)) ^(2) = (v_(i)) ^(2) + 2.a.d`

`(-3.66)^(2) = 3^(2) + 2.5.88.d`

d =
`(4.3956)/(11.76)`

d = 0.3738

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