Question

An 11.0 kg object traveling south with momentum of 50.0 kg-m/s collides and sticks to a 9.50 kg object traveling directly west with momentum 75.0 kg-m/s. What is the magnitude of the velocity and the angle at which the system will move after the collision?

Answer 1

Magnitude of velocity = 6.098 m/s and the angle at which system moves after collision is
`49.185^(o)`

Step-by-step explanation:

Given:

mass of object 1 m1 = 11.0 kg

initial momentum of object 1, p1 = 50 kg-m/s

mass of object 2 m2 = 9.5 kg

initial momentum of object 2, p2 = 75 kg-m/s

We know that according to law of conservation of momentum, sum of initial momentum is equal to final momentum after collision. Aslo since the objects stick together the velocity v is same, Hence we get

p1 + p2 = (m1 + m2) x v

substituting the known values

50 + 75 = (11 + 9.5) x v

125 = 20.5 x v

v = 6.098 m/s

To find the direction of velocity:

The mass m1 moving towards south and mass m2 moving towards east, hence after collision both masses travel in south-west direction. In order to find angle Θ (refer image), we use

tan Θ =
`(p_(1,collison) )/(p_(2,collision) )` where,
`p_(1, collision)` and
`p_(2, collision)` are momentum values after collision.

`p_(1, collision)` = m1 x v = 11 x 6.098 = 67.078 kg.m/s

`p_(2, collision)` = m2 x v = 9.5 x 6.098 = 57.931 kg m/s

Hence,

tan Θ =
`(67.078)/(57.931)`

Θ =
`tan^(-1)` (1.1579)

1. Θ =
   `49.185^(o)`