Answer:
Approximately [tex]0.29\; {\rm m \cdot s^{-1}}[/tex].
Explanation:
Make use of the fact that total momentum is conserved in collisions.
The momentum of an object of mass [tex]m[/tex] and velocity [tex]v[/tex] is [tex]p = m\, v[/tex].
The momentum of the two trolleys before the collision would be:
- [tex]4\; {\rm kg} \times 0.5\; {\rm m \cdot s^{-1}} = 2\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
- [tex]3\; {\rm kg} \times 0\; {\rm m\cdot s^{-1}} = 0\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Thus, the total momentum of the two trolleys right before the collision would be [tex]2\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Since the two trolleys are stuck to one another after the collision, they could modelled as one big trolley of mass [tex]m = 3\; {\rm kg} + 4\; {\rm kg} = 7\; {\rm kg}[/tex].
The momentum of the two trolleys, combined, is conserved during the collision. Thus, the total momentum of the new trolley of mass [tex]m = 7\; {\rm kg}[/tex] would continue to be [tex]v = 2\; {\rm kg \cdot m \cdot s^{-1}}[/tex] shortly after the collision.
Rearrange the equation [tex]p = m\, v[/tex] to find the velocity of the two trolleys combined:
[tex]\begin{aligned}v &= \frac{p}{m} \\ &= \frac{2\; {\rm kg \cdot m \cdot s^{-1}}}{7\; {\rm kg}} \\ &\approx 0.29\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
