Question

A particle is travelling at 2300 m/s in the x-direction. It collides with another particle travelling in the opposite direction at 100 m/s. After the collision, the first particle travels at 2000 m/s, at 45 degrees from the x-direction. What is the horizontal and vertical velocities of the second particle after the collision? Both particles have same mass (you actually don’t need to know the number).

Answer 1

`v'_(x) = 785.786 m/s`

`v'_(y) = 1414.214 m/s`(decreasing)

Given:

`u_(x)` = 2300 m/s

`u'_(x)` = 100 m/s

`v_(x)` = 2000 m/s

Angle made with the horizontal,
`\theta =  45^(\circ)`

The horizontal component of velocity is on the X-axis whereas the vertical one is on the Y-axis

Now, by the law of conservation of momentum for horizontal axis:

`mu_(x) + m'u'_(x) = mv_(x) + m'v'_(x)`

`2300 + (- 100) = 2000cos45^(\circ) + v'_(x)`

(The mass of the particles is same)

`v'_(x) = 2200 - 1414.214 = 785.786 m/s`

Now, by the law of conservation of momentum for vertical axis:

`mu_(y) + m'u'_(y) = mv_(y) + m'v'_(y)`

(The mass of the particles is same)

`u_(y) + u'_(y) = v_(y) + v'_(y)`

`0 = v_(y) + v'_(y)`

(since, initially, there's no vertical component of velocity)

`v'_(y) = - 2000sin45^(\circ) = 1414.214 m/s`(decreasing)

velocity, v =
`\sqrt{v^(2)_(x) + v^(2)_(y)}`

v =
`\sqrt{(785.786)^(2) + (1414.214)^(2)} = 1617.856 m/s`