Question

A 6.3 kg sphere with a speed of 6.7 m/s collides with a 0.75-kg ball, which deflects at an angle of 65º (to the direction the sphere was originally moving) and with a speed of 11.5 m/s. Find the final velocity (magnitude and direction) of the sphere.

Answer 1

According to law of conservation of momentum, along x- direction;

   $\small m_{s_{i}}v_{s_{i}}+m_{b_{i}}v_{b_{i}}=m_{s_{f}}v_{s_{f}}+m_{b_{f}}v_{b_{f}}$

$\small \Rightarrow (6.3)(6.7)+(0.75)(0)=(6.3)v_{s_{f}}cos\theta +(0.75)(11.5)(cos65^{\circ})$

$\small \Rightarrow v_{s_{f}}cos\theta =6.12$

Now along y- direction;

$\small m_{s_{i}}v_{s_{i}}+m_{b_{i}}v_{b_{i}}=m_{s_{f}}v_{s_{f}}+m_{b_{f}}v_{b_{f}}$

$\small \Rightarrow (6.3)(0)+(0.75)(0)=-(6.3)v_{s_{f}}sin\theta +(0.75)(11.5)(sin65^{\circ})$

$\small \Rightarrow v_{s_{f}}sin\theta =1.24$

So magnitude of final velocity of the sphere can be calculated as;

   $\small ( v_{s_{f}}sin\theta )^{2}+( v_{s_{f}}cos\theta )^{2}=(6.12)^{2}+(1.24)^{2}$

   $\small \Rightarrow v_{s_{f}}=6.24\: m/s$

and direction can be calculated as;

$\small \frac{ v_{s_{f}}sin\theta }{ v_{s_{f}}cos\theta }=\frac{6.12}{1.24}$

$\small \Rightarrow \theta =78.546^{\circ}$