Q.A child sits stationary at one end of a long trolley moving uniformly with a speed $V$ on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

The gun-shell system starts at rest, so by conservation of momentum the gun recoils with momentum equal and opposite to the shell's momentum.

Given:

Mass of shell, m1 = 200 g = 0.2 kg

Mass of gun, m2 = 100 kg

Muzzle speed of shell, v1 = 80 m/s

Before firing, the gun+shell system is at rest, so total initial momentum = 0. By conservation of momentum (no external horizontal force):

m1 v1 = m2 v2 (in magnitude, opposite directions)

0.2 x 80 = 100 x v2

16 = 100 v2

v2 = 0.16 m/s = 16 cm/s

True. The recoil of a gun is a classic real-world demonstration of conservation of linear momentum.

Before firing, the gun and the bullet inside it are both at rest, so the total momentum of the (gun + bullet) system is zero.

When the gun is fired, the bullet is pushed forward and gains a forward momentum m(v). Since no external horizontal force acts on the (gun + bullet) system during firing (the explosive force is internal), the total momentum of the system must remain zero, exactly as it was before firing.

So the gun must acquire an equal and opposite momentum, i.e. Mv(gun) = -m v(bullet), which means the gun recoils backward. This is also consistent with Newton's third law: the force the expanding gases exert on the bullet (forward) has an equal and opposite reaction force on the gun (backward).

Rocket propulsion is based on the conservation of linear momentum.

A rocket burns fuel and expels hot gases downward/backward with large momentum. Since no external force acts (ideally) on the rocket-gas system, total momentum is conserved: the backward momentum of the ejected gases is balanced by an equal forward momentum gained by the rocket, thrusting it forward.

Conservation of momentum for the gun-bullet system (initially at rest) requires the gun's recoil momentum to exactly cancel the bullet's forward momentum: m_gun v_gun = -m_bullet v_bullet, giving v_gun = -4 m/s.

Before firing, the gun+bullet system is at rest, so its total momentum is zero.

By conservation of momentum, the total momentum immediately after firing must also be zero:

m_bullet v_bullet + M_gun v_gun = 0

Given: m_bullet = 40 g = 0.04 kg, v_bullet = +800 m/s (taking the bullet's direction as positive), M_gun = 8 kg.

(0.04)(800) + (8) v_gun = 0

32 + 8 v_gun = 0

v_gun = -32/8 = -4 m/s

The negative sign shows the gun recoils in the direction OPPOSITE to the bullet, with speed 4 m/s.

Momentum p = mv is a vector because it is the product of a scalar (mass) and a vector (velocity); it therefore has both magnitude and direction, the direction being the same as that of the velocity.

Momentum is defined as p = mv, where m (mass) is a positive scalar and v (velocity) is a vector. Multiplying a vector by a positive scalar only scales its magnitude — it never removes its directionality. Hence momentum inherits a direction, always the same as the direction of the object's velocity, along with a magnitude |p| = m|v|.

Rocket propulsion is a direct application of the law of conservation of linear momentum.

Initially the rocket (with its fuel) is at rest, so the total momentum of the rocket-plus-fuel system is zero. As fuel burns, gas is ejected backward through the nozzle at very high velocity. Since no external horizontal force acts on the rocket-gas system, its total momentum must remain conserved.

If a small mass dm of gas is ejected backward with velocity v (relative to the rocket), it carries backward momentum. For the total momentum of the system to stay constant (zero, if starting from rest), the rocket must gain an equal and opposite forward momentum. This gives rise to a forward force called thrust:

Thrust F = v (dm/dt)

where dm/dt is the rate at which fuel mass is ejected and v is the exhaust speed relative to the rocket. This is exactly Newton's third law in action: the rocket pushes the gas backward, and the gas pushes the rocket forward with an equal and opposite force.

A rocket has no external horizontal force acting on the rocket-plus-exhaust system, so total linear momentum is conserved as fuel is ejected.

In rocket propulsion, hot gases are ejected backward at high speed. Since the rocket and the ejected gas together form an isolated system (ignoring gravity/air resistance for the basic principle), the total linear momentum of the system remains constant.

As the gas gains backward momentum, the rocket gains an equal and opposite (forward) momentum, which is the thrust that propels it -- a direct application of Newton's third law combined with conservation of linear momentum.

By conservation of momentum (initial total momentum = 0, since both gun and bullet start at rest), the gun recoils with a velocity that makes its momentum exactly cancel the bullet's forward momentum, giving 1 m/s.

Before firing, both the gun and bullet are at rest, so the total momentum of the system is zero. Since the force of firing is entirely internal to the gun-bullet system, total momentum is conserved:

m_bullet v_bullet + m_gun (-v_gun) = 0

(the gun's recoil velocity is taken as negative, i.e. opposite direction to the bullet)

m_bullet v_bullet = m_gun v_gun

Given: m_bullet = 0.2 kg, v_bullet = 5 m/s, m_gun = 1 kg

0.2 x 5 = 1 x v_gun

1.0 = v_gun

v_gun = 1.0 m/s

Momentum = mass x velocity = M x (L T^-1) = M L T^-1.

Linear momentum is defined as p = mv, the product of mass and velocity. Mass has dimension [M] and velocity has dimension [L T^-1], so momentum has dimensional formula [M L T^-1], matching option (e).

The statement is True: a rocket's forward thrust arises purely from conservation of linear momentum as it ejects mass (exhaust gases) backward.

A rocket carries its own fuel and oxidizer, and propels itself by burning fuel and ejecting the hot exhaust gases backward at very high speed. Before ignition, the total momentum of the rocket + fuel system is zero (or constant, if already moving). As gas is ejected backward with momentum, by the law of conservation of linear momentum (no external force needed for this internal process), the rocket itself must gain an equal and opposite momentum forward. This is exactly analogous to the recoil of a gun.

No air or any external medium is needed for this thrust (unlike, say, a propeller aircraft) — this is why rockets work even in the vacuum of space, since the momentum balance is entirely internal to the rocket-exhaust system.

Rocket propulsion is a direct consequence of conservation of linear momentum: gases are expelled backward, so the rocket is pushed forward.

A rocket works by burning fuel and ejecting the resulting hot gases at very high speed out of its rear nozzle. Before ignition, the rocket + fuel system is at rest (or moving with some velocity), with a certain total momentum. As fuel burns, exhaust gases of small mass but very high backward velocity are continuously ejected.

By the law of conservation of momentum (in the absence of external forces, or considering the closed rocket+exhaust system), the TOTAL momentum of the system (rocket + ejected gases) must remain constant. Since the ejected gas carries away momentum in the backward direction, the rocket itself must gain an equal and opposite (forward) momentum to keep the total unchanged. This is exactly what propels the rocket forward -- the reaction to the backward-ejected exhaust.

(This is also consistent with Newton's third law -- the rocket pushes the exhaust gas backward, and the gas pushes the rocket forward with an equal and opposite force -- but the underlying conserved quantity being used to derive the rocket equation and its thrust is momentum.)