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An object of mass $1 \,kg$ travelling in a straight line with a velocity of $10\, m\, s^{-1}$ collides with, and sticks to, a stationary wooden block of mass $5\, kg$. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
$12\, kg\, m\, s^{ - 1}$ , $5/6\, m/s$
$15\, kg\, m\, s^{ - 1}$ , $4/3\, m/s$
$20\, kg\, m\, s^{ - 1}$ , $3/5\, m/s$
$10\, kg\, m\, s^{ - 1}$ , $5/3\, m/s$
Solution
Mass of the object, $ m_1 = 1\, kg$
Velocity of the object before collision, $v_1 = 10\, m/s$
Mass of the stationary wooden block, $m_2 = 5\, kg$
Velocity of the wooden block before collision, $v_2 = 0\, m/s$
$\therefore $ Total momentum before collision $= m_1 v_1 + m_2 v_2$
$= 1 (10) + 5 (0) = 10\, kg\, m\, s^{ – 1}$
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system $= m_1 + m_2$
Velocity of the combined object $= v$
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
$m_1 v_1+ m_2 v_2 = (m_1+ m_2) v$
$1 (10) + 5 (0) = (1 + 5) v$
$v=\frac{10}{6}=\frac{5}{3} \,m / s$
The total momentum after collision $= 10$ and velocity of combined object is $5/3\, m/s$.
