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

Answer:

$v'_{x} = 785.786 m/s$

$v'_{y} = 1414.214 m/s$(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 = $\sqrt{(785.786)^{2} + (1414.214)^{2}} = 1617.856 m/s$