Motion of Connected Bodies - NEET Physics Chapterwise MCQs & PYQs

NEET Motion of Connected Bodies MCQs & PYQs

Question 1:

moderate

Two blocks are in contact on a frictionless table. One has mass m and the other 2 m. A force F is applied on 2 m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of force of contact between the two blocks will be:

Acceleration of the blocks in both the case will be F/3m.

N1= m.a

N2= 2m.a

so N1/N2= 1/2

Question 2:

moderate

Two forces of 6N and 3N are acting on the two blocks of 2kg and 1 kg kept on frictionless floor. What is the force exerted on 2kg block by 1 kg block?

Net force is 6N-3N= 3N

Accleration of the system is 3N/3Kg = 1 m/s^2

Net Force on 1kg object is N1 - 3= 1*1 so, N1 = 4N

 

Question 3:

moderate

A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 ms-2. The tension T1 and T2 will be respectively: (use g = 9.8m/s2)

Taking both the objects as one system

T1- 20g=20a ⇒ T1= 20(g+a)=20*12= 240 N

Now taking lower object as a different system

T2-8g=8a ⇒T2=8(g+a)=8*12= 96N

Question 4:

moderate

Two blocks of masses 1 kg and 2 kg are connected with massless spring as shown. If the acceleration of 1 kg block is 1 m/s² towards right then the acceleration of 2 kg block is: 

NLM Question NEET Physics

Spring force acting on 1 kg block is F = m.a = 1 N

NLM Question NEET Physics

Same spring force will act on 2 kg block in opposite direction. Thus net force acting on 2 kg block is 5N - 1 N = 4N

Acceleration of the block will be 4 N/2 kg = 2 m/s²

Question 5:

easy

Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is
applied on the 4 kg block, then the contact force between A and B is

connected body nlm

The total mass of the system:

M=4+2+1=7 kgM = 4 + 2 + 1 = 7 \text{ kg}

The acceleration of the system:

a=Total ForceTotal Mass=147=2 m/s2a = \frac{\text{Total Force}}{\text{Total Mass}} = \frac{14}{7} = 2 \text{ m/s}^2

Now, considering block B and C together (mass =

2+1=32 + 1 = 3

kg), the force required to accelerate them:

FAB=(3×2)=6 NF_{AB} = (3 \times 2) = 6 \text{ N}

Thus, the contact force between A and B is 6 N.

Question 6:

easy

A spring of length \(L\) and spring constant \(K\) is cut into two parts of length \(\frac{L}{3}\) and \(\frac{2L}{3}\), then the spring constant of each part will be

The spring constant is inversely proportional to the length of the spring: \(k \propto 1/l\). For lengths \(L_1 = L/3\) and \(L_2 = 2L/3\), the spring constants are \(k_1 = 3K\) and \(k_2 = 1.5K = 3K/2\).