(a) Both vehicles experience equal forces of action and reaction.
(b) Momentum change for both vehicles is same.
(c) Car experiences greater acceleration due to its smaller mass.
(d) Due to its smaller inertia or opposite to the force exerted on it.

Here     m = 1,200 kg
Initial velocity,  \(v = 90 \ km/h = 90 \times \frac {5}{18} \ m \ s^{-1}=25 \ m s ^{-1}\)
Final velocity,    \(u = 18 km/h = 18 \times \frac {5}{18} = 5m\quad s^{-1}\)
Time,    t  =  4 s
Accleration,   \(a = \frac {v-u}{t}=\frac {5-25}{4}=-5 m \ s^{-2}\)
magnitude of accleration  \(= 5 m \ s^{-2}.\)
Change in momentum \(= m \left( v- u \right)=1,200 \left(5-25\right)\)
 \(= - 24,000 \ kq \ m \ s^{-1}.\)
Magnitude of change in momentum \(= 24,000 \ kq \ m \ s^{-1}.\)
Magnitude of force \(= \frac {Change \ in \ momentum}{time \ taken}=\frac {24,000}{4}=6,000 \ N.\)

(a) The above table shows that the motion is accelerated and acceleration is increasing uniformly with time.
(b) As the acceleration is increasing uniformly, the force acting on the body is also increasing uniformly with time.

Here,   \(m_1 = 1 \ kg,\)    \(u_2= 10 \ ms^{-1},\)   \(m_2 = 5 \ kg,\)   \(u_2=0\)
Let v be the velocity of the combined object after the collision
Total momentum just before the impact
 \(=m_1u_1+m_2u_2 = 1 \times 10 +5 \times 0 = 10\quad kg \quad ms^{-1}\)
Total momentum just after the impact
 \(= (m_1+m_2)v=(1+5)v=6 v \quad kg m s^{-1}\)
By conservation of momentum,
6v = 10
or    \(v = \frac {10}{6}=\frac {5}{3} m s^{-1}\)
\(\therefore\)   Total momentum just after the impact
 \(= 6 \times \frac {5}{3}=10 \quad m s^{-1}.\)

Here,   \({ m }_{ 1 }={ m }_{ 2 }=1.5kg,\ { u }_{ 1 }=2.5{ ms }^{ -1 },\ { u }_{ 2 }=-2.5{ ms }^{ -1 }\) 
Let v be the velocity of the combined object after the collision. By conservation of momentum,
Total momenta after collision = Total momenta before collision
 \(\left( { m }_{ 1 }+{ m }_{ 2 } \right) v={ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }\)
\(\left( 1.5+1.5 \right) v=1.5\times 2.5+1.5\times \left( -2.5 \right) \)
 \(3.0\ v=0\)
 \(v=0\ { ms }^{ -1 }.\)

Total mass of engine and 5 wagons, \(m = 8,500 + 5 \times 2,000 = 18,000 kg\)
(a) The net accelerating force,
 F = Engine force - friction force = 40,000 - 5,000
= 35,000 N.
(b) The acceleration of the train,
 \(a = \frac {F}{m}=\frac {35,000}{18,000}=\frac {35}{18}=1.94 ms^{-2}\) 

For boy: \({ m }_{ 1 }=60\ kg,\)   \({ u }_{ 1 }=3m\ { s }^{ -1 }\) \({ v }_{ 1 }=v\) (say)
For trolley: \(\\ { m }_{ 2 }=140\ kg,{ u }_{ 2 }=1.5\ { ms }^{ -1 },{ v }_{ 2 }=v \)
By conservation of momentum
\({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }=\left( { m }_{ 1 }+{ m }_{ 2 } \right) v\)
\(60\times 3+140\times 1.5=\left( 60+140 \right) v\)
or   \(v=\frac { 180+210 }{ 200 } =\frac { 390 }{ 200 } =1.95{ ms }^{ -1 }.\)

Here,  \(m_1 = 2\) kg,  \(u_1 = 10 m{s}^{-1}\)   \(v_1 = 1 {m}{s}^{-1}\)
and \(m_2 = 5kg,\)   \(v_2 = 0,\)  \(v_2 =?\)
By conservation of momentum,
 \({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }\)
\(2\times 10+5\times 0=5\times 1+5{ v }_{ 2 }\)
\(\therefore\)   \({ u }_{ 2 }=\frac { 15 }{ 5 } =3{ ms }^{ -1 }.\)

Let the final velocity of the bag along with the bullet embedded in it be u.
\(\therefore\)  For bullet,   \(m_1=20\) g = 0.02 kg,  \(u_1 = 500 m/s,\)   \(v_1=v\)
 For block,    \(m_2= 1\) kg,    \(u_2=0,\)    \(v_2=v\)
According to the law of conservation of momentum,
Total momentum before collision  = Total momentum after collision
\({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }\)
or \({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }=\left( { m }_{ 1 }+{ m }_{ 2 } \right) v\)    \([\because v_1=v_2=v]\)
or   \(v=\frac { { m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
\(=\frac { 0.02\times 500+1\times 0 }{ 0.02+1 } =\frac { 10 }{ 1.02 } =\frac { 1000 }{ 102 } =9.87\quad m/s\)

Mass of bullet, \(m_1=50 \ g = 0.05 kg\)
Mass of rifle,   \(m_2=4\) kg
Initial velocity of bullet, \(u_1=0\)
Initial velocity of rifle, \(u_2=0\)
Final velocity of bullet, \(v_1=35 ms^{-1}\)
Final velocity of rifle, \(v_2 = ?\)
According to the law of conservation of momentum,
total momenta after the fire = Total momenta before the fire
 \(m_1v_1+m_2v_2=m_1u_1+m_2u_2\)
 \(0.05 \times 35 + 4v_2=0+0\)
\({ v }_{ 2 }=-\frac { 0.05\times 35 }{ 4 } =-\frac { 7 }{ 16 } =-0.44{ ms }^{ -1 }\)
The negative sign indicates that the direction in which the rifle would recoil is opposite to that of the bullet.    

Law of conservation of momentum. This law states that if a number of bodies are interacting with each other (i.e., exerting forces on each other), their total momentum remains conserved before and after the interaction, provided there is no external force acting on them.
Derivation from Newton's second law of motion.  Let  \(p_1\)   and   \(p_2\)   represent the sum of momentum of a group of objects before and after the collision, respectively. Let t be the time elapsed during the collision.
According to Newton's Second law of motion,
 External force = Rate of change of momentum
or     \(F = \frac {p_2-p_1}{t}\)
If there is no external force, that is F = 0, then
\(\frac {p_2-{p}_{1}}{t}=0\) or \(p_2 = p_1\)   
Hence in the absence of an external force, the total momentum of a group of objects remains unchanged or conserved during the collision. This is the law of conservation of momentum.

(i) m = 5 kg,  u = 20 m/s, v = 0.20 m/s
Change in momentum = m (v-u) = 5 (0.20 - 20) kg m/s = - 99 kg m/s
The negative sign shown that the momentum decreases.
(ii) m = 5 kg,  u = 30 m/s,v = 40 m/s
Chnage in momentum = m (v - u) = 5 (40 - 30) = 50 kg m/s
Here, the momentum of the body increases.

Mass and inertia. The mass of a body is a measure of its inertia. The larger the mass of a body, the larger is inertia or opposite offered by a body to change its state of motion. This can be understood from the following examples:
(i) If we kick a football, it flies a long way. if we kick a stone of the same size, it hardly moves instead, we may get an injury in our leg while hitting the stone. The stone has more inertia than the football.
(ii) we may cause a bicycle to pick up a large velocity by applying a certain force. But the same force will produce a negligible change in the motion of a train. This is because, in comparison to the bicycle, the train has more to oppose any change in its state of motion because of its larger mass. in order words, the train has more inertia than the bicycle.
(iii) The SI unit of mass and inertia is kilogram (kg).

Unbalanced forces. If the resultant of the several forces acting on a body is not zero, the forces are said to be unbalanced forces. Unbalanced forces produce a change in the state of rest or uniform motion of a body.
Examples:
(i) In a tug-of-war, when one of the two teams pulls the rope with a larger force, it is able to pull the weaker team towards it. Here teh two forces are not balanced. Therefore, is results in the motion of the weaker team towards the larger force along the rope.
(ii) When we stop pedaling a bicycle, it begins to slow down the road has small cracks and bumps. The bicycle has to overcome these imperfections, which slow down it. Forces which slow down the moving objects in this way recalled forces of friction.
so we can say that objects continue to move, with the same velocity unless acted upon by unbalanced forces.