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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 $2.5\, m \,s^{-1}$ before the collision during which they stick together. What will be the velocity(in $m/s$) of the combined object after collision?
$3$
$1$
$2$
$0$
Solution
Mass of one of the objects, $m_1 = 1.5\, kg$
Mass of the other object, $m_2 = 1.5\, kg$
Velocity of $m_1$ before collision, $v_1 = 2.5\, m/s$
Velocity of $m_2$, moving in opposite direction before collision, $v_2 = -2.5\, m/s$
(Negative sign arises because mass $m_2$ is moving in an opposite direction)
After collision, the two objects stick together.
Total mass of the combined object $= m_1 + m_2$
Velocity of the combined object $= v$
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
$m_1v_1 + m_2 v_2 = (m_1 + m_2) v$
$1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5) \,v$
$3.75 – 3.75 = 3\, v$
$v = 0$
Hence, the velocity of the combined object after collision is $0\, m/s$.
Similar Questions
The following is the distance-time table of an object in motion:
Time in seconds | Distance in metres |
$0$ | $0$ |
$1$ | $1$ |
$2$ | $8$ |
$3$ | $27$ |
$4$ | $64$ |
$5$ | $125$ |
$6$ | $216$ |
$7$ | $343$ |
$ (a)$ What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
$(b) $ What do you infer about the forces acting on the object?