Question

a) Find the Velocity of object B immediately after the collision b) Find the speed of object B immediately after the collision c) Find the direction of object B immediately after the collision,giving the answer as an angle from the positive x axis in degrees to 1 decimal placed) Find the direction of object B immediately after the collision,giving the answer as bearing in degrees to 1 decimal place

Answer 1

Given:

the mass of the object A is

$m_A=2\text{ kg}$

The mass of the object B is

$m_B=4\text{ kg}$

The velocity of the object A before the collision is

$u_A=-9i+11j\text{ m/s}$

The velocity of the object B before the collision is

$u_B=i-j\text{ m/s}$

The velocity of the object A after the collision is

$v_A=-i+5j\text{ m/s}$

Required: the multiple parts to be solved

Explanation:

to solve this problem, we will use momentum conservation.

according to the momentum conservation.

momentum before the collision = momentum after the collision.

we will conserve the momentum in the respective x and y directions.

the object A is the velocity in x and y direction is

$\begin{gathered} u_{Ax}=-9\text{ m/s} \\ u_{Ay}=11\text{ m/s} \end{gathered}$

The object B velocity in the x and y direction is

$\begin{gathered} u_{Bx}=\text{ 1 m/s} \\ u_{By}=-1\text{ m/s} \end{gathered}$

(a)

momentum conservation in x direction is

momentum before the collision = momentum after the collision

$\begin{gathered} P_{ix}=P_{fx} \\ m_Au_{Ax}+m_Bu_{Bx}=m_Av_{Ax}+m_Bv_{Bx} \end{gathered}$

Plugging all the values in the above relation, we get:

$\begin{gathered} 2\text{ kg}\times-9\text{ m/s }+4\text{ kg}\times\text{1 m/s }=2\text{ kg}\times-1\text{ m/s}+4\text{ kg}\times v_{Bx} \\ v_{Bx}=-\text{ 3 m/s} \\ \end{gathered}$

now apply momentum conservation in y direction,

$\begin{gathered} P_iy=P_{fy} \\ m_Au_{Ay}+m_Bu_{By}=m_Av_{Ay}+m_Bv_{BY} \end{gathered}$

plugging all the values in the above relation, we get;

$\begin{gathered} 2\text{ kg}\times11\text{ m/s}+4\text{ kg}\times-1\text{ m/s }=2\text{ kg}\times\text{5 m/s}+4\text{ kg}\times v_{By} \\ v_{By}=2\text{ m/s} \end{gathered}$

the velocity of the object B after the collision is

$\begin{gathered} v=v_{Bx}+v_{By} \\ v=-3i+2j\text{ m/s} \end{gathered}$

Thus, the velocity of the object B is

$v=-3\imaginaryI+2j\text{ m/s}$

(b)

the speed of the object after the collision can be calculated as

$\begin{gathered} speed=\sqrt{v^2_{x^{\text{ }}}+v_y^2} \\ speed=\sqrt{-3^2+2^2} \\ speed=3.60\text{ m/s} \end{gathered}$

Thus, the speed of the object B is 3.60 m/s.

(c)

The direction of the object B can be found as,

the velocity of the object B after the collision is

'

$\begin{gathered} v_{Bx}=\text{ -3 m/s} \\ v_{By}=2\text{ m/s} \end{gathered}$

direction is given by

$\tan\theta=\frac{v_{By}}{v_{Bx}}$

Plugging all the values in the above relation, we get:

$\begin{gathered} \tan\theta=\frac{2\text{ m/s}}{-3\text{ m/s}} \\ \theta=\tan^{-1}(-0.666) \\ \theta=123.24\text{ degree from the positive x axis} \end{gathered}$

Thus, the direction is 123.24 degree.

(d)

f

The direction calculated as

$\begin{gathered} \tan\theta=\frac{2\text{ m/s}}{-3\text{ m/s}} \\ \tan\theta=-0.66 \\ \theta=33.24\text{ degree.} \end{gathered}$

Thus, the direction is 33.24 degree.