31 Year NEET Previous Year Questions: Laws of Motion - 1 - NEET MCQ

Velocity just before striking the ground

If it reaches the same height, speed remains same after collision only the direction changes.

Speed = 10 m/s; radius r = 10 m
Angle made by the wire with the vertical is:

From figure, F = 6 mg
As speed is constant, acceleration a = 0
∴ 6 mg = 6ma = 0 
F = 6 mg
T = 5 mg , T' = 3 mg T" = 0
F_net on block of mass 2 m = T – T' – 2 mg = (5 - 3 - 2)mg = 0
Alternate Method:
 v  = constant  so, a = 0, Hence, F_net = ma = 0

Momentum,  

(v² = u² + 2gh; Here u = 0)

When a stone hits the ground momentum:

When the same stone dropped from 2h (100% of initial) then momentum:

which is changed by 41% of the initial.

or

For banking, tanθ = V²/ Rg 
tan 45° = V² / 90*10 = 1
V = 30 m/s

Frictional force on the box, f = μmg
∴  Acceleration in the box, μg = 5 ms^-2

v² = u² + 2as
0 = 2² + 2 * 5 * s
s = -2 / 5 w.r.t belt

Therefore distance = 0.4 m

A Body of mass =m

velocity =V(i^)

Knock out (Bounce velocity) =V(−i^)

Change in momentum =m(v)−(m)(−v)
=2MV
The rate of change of momentum is Force
Thus dp/dt​​= force.
force × time = Impulse
Thus Impulse Experience by body =2mv.

Total mass = (60 + 940) kg = 1000 kg
The tension in the string is given as follows:

T=m(g+a)

By substituting the values of mass and acceleration given in the question we get

T=1000×(10+1)

=11000N

Hence, the tension in the string is11000N

The gravitational force in the downward direction is 20000 N and the tension force in the upward direction is 28000 N.

By newton's second law of motion, ΣF=ma

Net force, F = T – mg
ma = T – mg
2000 * a = 28000 – 20000 = 8000

Work done = 300J, Mass = 2kg, Height = 10m, g = 10m/s²
We know that the work done = f.d
300 = (mgsinθ + f) * (h / sinθ)

Work done against the friction:
f.(h / sinθ) = 300 - mgsinθ * (h / sinθ)
300 - mgh = 300 - 2x10x10 = 100 J

Friction is the retarding force for the block F = ma = μR = μmg
From the first equation of motion:

v = u – at

0 = v−μgt

t = v​/μg

Accelaration, a = 1 m/s

m = mass of the system as a conveyor belt with sand drops at time t.

∵ v → constant
∴ 

According to Newton's second Law:

N = masinθ + mgcosθ     .....(1)
Also, mgsinθ = macosθ  .....(2)

From (1) & (2) we get, a = gtanθ

For smooth driving maximum speed of car v then: