Question

8. 20-kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0.500 m/s. At the origin 0, the bomb explodes into two fragments. Fragment 1 has a mass of 0.400 kg and a speed of 0.900 m/s along the negative y- axis. In the figure, the angle , is made by the velocity vector of fragment 2 and the x-axis. Find the momentum vector of fragment 2 after the collision using conservation of momentum (not a graphical or trig method) and find its speed and direction too. 0.80 kg 0.50 m/s 1.2 kg Before After ' Alter 0.40 kg 0.90 m/s

Answer 1

conservation of momentum before and after impact

$m\vec{v_{i}}=m_{1}\vec{v}_{1,final}+m_{2}\vec{v}_{2, final}$

along x- axis :

$(1.2kg)(0.5m/s)=(0.4kg)(0)+(0.8)v_{2}cos \theta$$\Rightarrow v_{2}cos \theta = 0.75m/s$

along y axis :

$0 = (0.4kg)(-0.9m/s)+(0.8kg)v_{2}sin \theta$$\Rightarrow v_{2}sin \theta = 0.45m/s$

momentum vector after collision

$\vec{v}=v_{2}cos \theta \hat{i}+ v_{2}sin \theta \hat{j}= 0.75m/s\hat{i}+0.45m/s\hat{j}$

$tan \theta = \frac{0.45}{0.75}=0.6\Rightarrow \theta = 31^{\circ}$

$|\vec{v}|=\sqrt{(0.75)^{2}+(0.45)^{2}}=0.87m/s$