Test: Newton's Second Law of Motion (NCERT)

Newton's second law of motion is
F = dp/dt = ma

The second law of motion is a local law which means that force `F` at a point in space (location of the particle) at a certain instant of time is related to a at that instant. Acceleration here and now is determind by the force here and now, not by any history of the motion of the particle.

The relation , can only be deduced from Newton's second law, if mass remains constant with time. If mass depends on time then this relation cannot be deduced.

If a large force F acts for a short time dt the impulse imparted I is
I = Fdt = dp/dt dt
I = dp = change in momentum

Tension, thrust, weight are all common forces in mechanics whereas impulse is not a force.
Impulse = Force x Time Duration

Here  y = ut + 1/2gt²
∴ v = dt/dy  = u + gt
Acceleration, a = dt/dv = g
So, the net force acting on the particle is, F = ma = mg

Here,m = 5kg, M = 5ms^-1, v- 10ms^-1, t = 10s, Using v = u + at

As F = ma
∴ F = (5kg)(0.5ms⁻²) = 2.5N

Here, u = 90  ms⁻¹, v = 0
m = 40gm = 4/1000kg = 0.04 kg, s = 60 cm = 0.6 m
Using v² − u²  = 2as
∴ (0)² − (90)²  = 2a × 0.6

-ve sign shows the retradation.
∴ The average resistive force exerted by block on the bullet is
F = m × a = (0.04kg)(6750ms⁻²) = 270N

Here m = 5kg
∴ u_y = 40 m s^-1 and F_y = -5 N
 
= 1 m s^-1 v_y = 0
As vy = uy + ayt or 0 = 40-1 x t
∴  t = 40s

Given: and a = 5ms⁻²
∴ 
Mass of the body is, m = F/a
= 10N / 5ms⁻²
= 2 kg

Here, F = -50 N (-ve sign for retardation) m=10 kg, u = 10 ms⁻¹, v = 0
As F = ma 
∴ a = 
Using v = u + at
∴ t = 
= 2 s

Mass of the particle, m = 0.4 kg
F = -8 N (minus sign for direction of force)
∴ Acceleration, a = F/m = 
= -20 m s^-2
The position of the body at any time t is given by

The position of the body at t = 0 is 0, therefore x₀ = 0

Position of the body at t = 25 s
Here, u = 10ms⁻¹, a = −20ms⁻², t = 25s
∴ x = 10  × 25 + 1/2(−20)(25)²
  = 250 − 6250 = −6000m

Force = d/dt(momemtum)
= d/dt(mv) = v(dm/dt) ⇒ 210 = 300(dm/dt)
 rate of combustion, dm/dt = 210/300 = 0.7 kgs^-1

The situation is as shown in the figure.

p_initial = mu,p_final = −mu
Impulse imparted to the ball = change in momentum = p_initial = mu,p_final
 = −mu − mu = −2mu

Change in momentum = 0.15 x 12 - (-0.15 x 12)
= 3.6 N s