Question.
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms–1 before the collision during which they stick together. What will be the velocity of the combined object after collision ?
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms–1 before the collision during which they stick together. What will be the velocity of the combined object after collision ?
Solution:
Let the objects be $\mathrm{A}$ and $\mathrm{B}$ moving from opposite directions in the same straight line.
$\therefore$ Initial momentum of $\mathrm{A}=\mathrm{m} \times \mathrm{v}$
$=1.5 \times 2.5=3.75 \mathrm{~kg} \mathrm{~ms}^{-1}$
Also, initial momentum of B
$=\mathrm{m} \times \mathrm{v}=1.5 \times(-2.5)=-3.75 \mathrm{~kg} \mathrm{~ms}^{-1}$
If v is the velocity of the objects after collision, then final momentum after collision = 3 × v
Using the law of conservation of momentum, final momentum of A and B = initial
momentum of A + initial momentum of B
or $3 \times v=3.75-3.75$
or $\quad 3 \times \mathrm{v}=0$ or $\mathrm{v}=\mathbf{0}$
Let the objects be $\mathrm{A}$ and $\mathrm{B}$ moving from opposite directions in the same straight line.
$\therefore$ Initial momentum of $\mathrm{A}=\mathrm{m} \times \mathrm{v}$
$=1.5 \times 2.5=3.75 \mathrm{~kg} \mathrm{~ms}^{-1}$
Also, initial momentum of B
$=\mathrm{m} \times \mathrm{v}=1.5 \times(-2.5)=-3.75 \mathrm{~kg} \mathrm{~ms}^{-1}$
If v is the velocity of the objects after collision, then final momentum after collision = 3 × v
Using the law of conservation of momentum, final momentum of A and B = initial
momentum of A + initial momentum of B
or $3 \times v=3.75-3.75$
or $\quad 3 \times \mathrm{v}=0$ or $\mathrm{v}=\mathbf{0}$