Question

A 0.300 kg ball, initially at rest on a horizontal, frictionless surface, is struck by a 0.200 kg ball moving initially along the x axis with a speed of 2.00m/s. After the collision, the 0.200 kg ball has a speed of 1.00 m/s at an agle of = 53.0 0 to the positive x axis. (a) Determine the velocity and direction of the 0.300-kg ball after the collision

Answer 1

Explanations & Calculations

• The linear momentum of the system of 2 balls is conserved due collision in a frictionless environment.
• After the collision the balls move away in a manner the total linear momentum is conserved.
• Therefore, consider the momentum along the moving direction

$\qquad\qquad \begin{aligned} \small m_1v_1+m_2v_2&= \small m_1V_1^1+m_2V_2^{11}\\ \small 0.2\times 2+0.3\times 0&= \small 0.2\times 1\cos 53\,+ 0.3V\cos \theta\\ \small V\cos\theta&= \small 0.932ms^{-1}\cdots(1)\\ \end{aligned}$

• That along the direction perpendicular to the moving direction

$\qquad\qquad \begin{aligned} \small 0.2\times 0+0.3\times 0 &= \small 0.2 \times 1\sin53\,+ 0.3\times (-V\sin\theta)\\ \small V\sin\theta &= \small 0.532 ms^{-1} \cdots(2) \end{aligned}$

• Now to calculate the magnitude of the velocity $\small \to \sqrt{(1)^2+(2)^2}$

$\qquad\qquad \begin{aligned} \small V &= \small \sqrt{0.932^2+0.532^2}\\ &= \small \bold{1.073\,ms^{-1}} \end{aligned}$

• To calculate the direction$\small \to (1)/(2)$

$\qquad\qquad \begin{aligned} \small \tan \theta &= \small \frac{0.532}{0.932}\\ \small \theta &= \small \tan^{-1} (0.57082)\\ &= \small \bold{29.72^0} \end{aligned}$