Test: Laws of Motion - 3 - NEET MCQ

The explanation is the question itself. Inertia is a property of which it resists its change of state of rest or state of motion, so as soon as the person jumps the lower part immediately comes to rest by sharing contact with the ground the upper body due to inertia of motion resists its change in state of motion.

Spitting or sneezing this is because you can not jump forward or backward or rolling or even run on the frictionless plane. So you have to throw anything out of your body by applying force Newton's third law follows here and you will escape out. 

As there is no net acceleration in horizontal or vertical direction, we can say that
P = 300N
And 2S = 2000N
Thus S = 1000N

Since no external force, law of conservation of momentum can be applied
Initial Momentum

Final momentum 
 
where v is the common velocity
Now we know that
Initial momentum = Final momentum
280 = 70v
V = 4m/s

Workdone by T = 0
∴ ∑T⋅V = 0
2T + T + 2TV = 0
⇒ − 3T = − 2TV (T & 2T are 180°)
∴ V = 3/2 m/s↑

l₁ = 2l₂ + l₃ = constant
dl₁/dt + 2dl₂/dt + dl₃/dt = 0
−5 + 2(−5) + dl/dt = 0
dl₃/dt = 15m/s
⇒ v_B = 15m/s↓

 

As the mass is at rest the resultant of forces acting on it are equal to zero so forces in vertical direction are 

T₁​sin30^o + T₂​sin60^o − mg = 0

similarly in horizontal direction

T₁​sin30^o − T₂​sin60^o = 0

solving above equations will give us 

T_1​+ √3​×√3​T_1​ = 100

T₁ ​= 25N

T₂​ = 25√3​N

Net Force = Force exerted by Cyclist - Frictional Force
Also, according to newton's second law
F_net = m.a
250 N - Frictional Force = 30x4
∴ Frictional Force = 250 - 120 N
= 130 N

Thus option D is correct

Let say end b has some velocity v in horizontal direction. Thus by constraint motion we get the component of velocities along the rod of both the ends must be equal thus we get, 
u.cos (90-q) = v.cos q
Thus we get v = u.tan q

As 250N is applied to move the cyclist and cycle, and lets say some frictional force f is acting upon it.
Thus we get 250 - f = 30 x 4 =120
Thus we get f = 130N

N = mgcosФ    
= 2 x 10 x cos30    
= 2 x 10 x √3 / 2  
= 10√3 N 

As the block is still at the equilibrium, we get 
T.cos q = mg
T.sin q = F
Thus we T = F / sin q

Let a be the acceleration of the system.
T₁​ = M_1​a  .....(1)
T₂ ​− T₁ ​= M₂​a  ....(2)
F − T₂​ = M₃​a  ......(3)
Adding (1), (2) and (3)  we get
(M_1​ + M₂​ + M₃​)a = F
or (10+20+30)a = 60
⇒ a = 1m/s²
Now , T_{2 ​}= (M₁​+M₂​)a
⇒ (10+20)(1) = 30N

Given,
Mass of block A & B = m
Mass of block C = M
Let,
Mass of A = m_A​ & Mass of B = m_B
Tension in string is = T
At equilibrium, T = m_A​g = m_B​g = mg
Weight of block C is = Mg 
Forces on block C,
2Tcosθ = Mgcosθ
⇒ Mg/2T = Mg/2mg
⇒ ​M/2m
If 0< θ <90° then 1> cosθ >0
1> M/2m >0
2m > M
Hence, 2m > M

• In all the given cases the cos component of the tension in the string would balance the weight of the block while the sine component will cancel themselves as they are in pair.
• The larger would be the angle the smaller would be the cos component but as its value is fixed i.e. is mg, we get the larger the angle the larger the tension and hence the more chances of breaking of the rope.

Momentum is rate of change of force according to the given graph initial and final forces are same ie, no change in force therefore b/w 0 and 8 seconds linear momentum is 0.

F_x = dp_x / dt = - 2sint
F_y = dp_y / dt = 2cost
So angle between F and P will be 90º  because we see that their dot product is zero.

• As the car is in free fall condition it's acceleration will be g downwards.
• Applying pseudo force mg in upward direction , it's weight acts downwards, hence, net force on stunt man is zero and it experience weightless ness.

• To know the force acting on the particle, we need to know the acceleration of the particle. In the velocity-time graph the slope of the curve gives the instantaneous acceleration. Let us find it for the given instant.
• At t=2 seconds
  The graph is a straight line with positive slope. It means the particle has a constant acceleration with magnitude =15/3 =5 m/s.
  So the force acting on it = mass x acceleration = 0.05 kg x 5 m/s
  = 0.25 N, it acts along the motion because it is positive.
• At t=4 seconds
  The graph is horizontal to time axis meaning thereby the velocity is constant and no acceleration. It can be understood in this way too that the slope which represents acceleration is zero. Since there is no acceleration at t=4 s. so there is no force acting on the particle at this instant, Force= zero.
• At t=6 seconds
  The graph shows that velocity is uniformly decreasing with the time and the acceleration which is represented by the slope is negative. From the graph value of acceleration = -15/3 =-5 m/s
  Force = mass x acceleration = 0.05 x -5 N =-0.25 N
• So the force acting on the particle is 0.25 N and negative sign shows that its direction is opposite to the motion.

The tension T₁ and T₂ will be 240 N, 96 N.
Mass of the body = m₁ = 12kg (Given)
Mass of the body = m₂ = 8kg (Given)
Acceleration of the string - 2.2m/s² (Given)
Considering positive direction upwards -  
Thus,
T1− (m₁+m₂)g = (m₁+m₂)a
T₁ = (m₁+m₂)g + (m₁+m₂)a
T₁= (12+8)(2.2)+(12+8)(9.8)
T₁ = 44 + 196
= 240 N
T₂− (m₂)g=(m₂)a
T₂=(m₂)g+(m₂)a
T₂= (8)(2.2)+(8)(9.8)
T₂ = 17.6 + 78.4  
= 96 N

When you are given one upward and one downward force on a rope of some mass, and you have to find the tension at the mid- point, then you can simply average the 2 forces. So, the correct answer is (100+70)/3 = 85N

The heavy body will fall due to accelaration due to gravity.The accelaration of the lighter body wil be 'g' in upward direction.Taking the equlibrium of the small body 

T - mg = ma

but a = g

Therefore

T - mg = mg

T = 2mg

reation at the pulley

2T = 4mg

When we make the free body diagram of two mass system in case 1, we get the equation
2mg - mg = (2m + m) a
Thus we get a = g/3
But in the case 2 the equations is as follows
2mg - mg = ma
Thus we get a - g here.

If we straighten the string horizontally hypothetically and make the free body diagram for the whole two block system we get that
M₁g.sinɑ - M₂g.sinβ = (M₁ + M₂)a
For some net acceleration of the system a
Such that a = M₁g.sinɑ - M₂g.sinβ / (M₁ + M₂)
Hence if we make a free body diagram of mass 1 only we get
M₁g.sinɑ - T = M₁a
Thus we get T = M₁ ( g.sinɑ - a )
= M₁( g.sinɑ - [M₁g.sinɑ - M₂g.sinβ / (M₁ + M₂) ] )
= M₁( M₂ (sinɑ + sinβ )f / (M₁ + M₂) )

Free Body Diagram:

F_net = m₂g - m₁g
F = (m₁ + m₂) a
m₂g - m₁g = (m₁ + m₂) a

For some time let us consider B and C to be one single block of mass 4kg. Now if we make the free body diagram for blocks A, B, C we get the net acceleration of the system by equation
4.g - 2.g = 6.a
Thus we get a = g/3
Now if we only make a F.B.D. of block C, we get
2.g - T = 2.a
Thus T = 2 (g - g/3)
= 4g/3
= 13.3 N

a = [(m_3​−m₂ )/(m₂​+m₃​) ​​]g  (m₃ > m₂)
T = [2m₂​m₃​g] / [m₂​+m₃] ​
​T′ = 2T = [4m₂​m₃​g​] / [m₂​+m₃​]
m₁​g = 4m₂​m_3​g​m / m₂​+m₃​
4/m₁​ = [1/m_2​​] + [1/m_3​]​​

For a small value of the applied force the force of friction f increases linearly upto the limiting friction. When it crosses the maximum point of static friction then the friction force due to kinetic friction does not change any more.