Question

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.

Answer 1

Hint: In extreme cases of inelastic collision, the masses stick together. In an inelastic collision the kinetic energy is not conserved, only the total energy is conserved. This also implies that the momentum also will be conserved in an inelastic collision. Almost all macroscopic collisions are inelastic collisions. Remember that after sticking together the composite body moves with one velocity ${v_f}$ and the mass will be the sum of both i.e. ${m_1} + {m_2}$.

Formula used:
$P = m \times v$, where $P$ denotes the momentum of the body, $m$ denotes the mass and $v$ denotes its velocity.
$M \times {v_f} = {m_1} \times {v_1} + {m_2} \times {v_2}$, where $M$ denotes the total or combined mass,${v_f}$ denotes the final velocity of the composite body, ${m_1}$ and ${m_2}$ denote the masses of the first and second body and ${v_1}$ and ${v_2}$ shows their velocities.

Complete step by step answer:
This is an inelastic collision; therefore there is no conservation of kinetic energy .But the total energy as well as the momentum of the system is conserved. We can use the conservation of momentum to solve this question. I.e. the sum of momentum of the two bodies which collided will be equal to the momentum of the composite body. This is represented by the equation$M \times {v_f} = {m_1} \times {v_1} + {m_2} \times {v_2}$, where $M$ denotes the total or combined mass,${v_f}$ denotes the final velocity of the composite body, ${m_1}$ and ${m_2}$ denote the masses of the first and second body and ${v_1}$ and ${v_2}$ shows their velocities. Remember $M \times {v_f}$ shows the momentum after impact and ${m_1} \times {v_1} + {m_2} \times {v_2}$ shows the momentum before impact.
It is given that ${m_1} = 1Kg$, ${v_1} = 10m/s$, ${m_2} = 5Kg$, ${v_2} = 0m/s$
(Note that the second object was stationary which implies its velocity is $0m/s$)
$M$ will be the total mass i.e. the sum of two masses $M = 6Kg$
Momentum before impact =${m_1} \times {v_1} + {m_2} \times {v_2}$
$ = (1 \times 10) + (5 \times 0)$
$ = 10kgm/s$
According to the conservation of linear momentum, the total momentum before and after impact will be the same.
Final momentum $ = 10kgm/s$
Using ${m_1} \times {v_1} + {m_2} \times {v_2}$$ = 10kgm/s$ and $M = 6Kg$ in $M \times {v_f} = {m_1} \times {v_1} + {m_2} \times {v_2}$
$6 \times {v_f} = 10$
${v_f} = \dfrac{{10}}{6} = \dfrac{5}{3}m/s$
This is the final velocity of the combined masses.

Note: Normally all ordinary collisions are considered to be inelastic. This is because for a collision to be considered as elastic its kinetic energy as well as its momentum/energy should be conserved. The kinetic energy is not conserved in ordinary collisions due to the action of internal friction. In most natural cases some of the kinetic energy is converted to other forms. For example in some cases, the kinetic energy is converted into vibrational energy of the atom, in some others as sound energy etc.