CBSE Class 9 Physics Force and Laws of Motion MCQs Set 09

Answer: (b) 1 : 2

Explanation:
1. Case 1: acceleration of both blocks = \( \frac{F}{3m} \). The contact force is the force that accelerates m: \( N_1 = m\times\frac{F}{3m}=\frac{F}{3} \).
2. Case 2: the contact force accelerates the 2m block: \( N_2 = 2m\times\frac{F}{3m}=\frac{2F}{3} \).
3. Ratio N₁ : N₂ = 1 : 2.

Teacher's Note:
1. Find the common acceleration of the whole system first.
2. The contact force equals mass of the block ahead times that acceleration.

Answer: (c) 4 N

Explanation:
1. Acceleration of the system: \( a=\frac{6-3}{2+1}=1\ \text{m/s}^2 \) towards the right.
2. For the 2 kg block: 6 − N = 2 × 1, so N = 4 N.

Teacher's Note:
1. Treat the two blocks as one body to get the acceleration.
2. Then apply F = ma to one block alone.

Answer: (c) \( -32\,\hat{i}-12\sqrt{3}\,\hat{j} \) N

Explanation:
1. Net force: 2 × 12 = 24 N, with components \( (-24\sin30^\circ,\,-24\cos30^\circ)=(-12,\,-12\sqrt3) \) N.
2. F₂ = net force − F₁ = \( (-12-20)\hat i-12\sqrt3\,\hat j=-32\hat i-12\sqrt3\,\hat j \) N.

Teacher's Note:
1. Resolve the net force into x and y components.
2. The unknown force equals the net force minus the known force.

Answer: (b) 0.098 m/s

Explanation:
1. Each bullet rebounds, so its momentum change is 2mv.
2. Force = (bullets per second) × 2mv = \( 100\times2\times0.005\,v = v \) newton.
3. This must balance the weight: \( v=0.01\times9.8=0.098 \) m/s.

Teacher's Note:
1. Rebound doubles the momentum change.
2. Equilibrium means force from bullets equals weight.

Answer: (a) 2.5 m/s²

Explanation:
1. Water of 1 kg per second moving at 5 m/s is brought to rest by the block, so force = (1)(5) = 5 N.
2. Acceleration = 5/2 = 2.5 m/s².

Teacher's Note:
1. Force = rate of change of momentum = (mass per second) × change in speed.
2. The block's mass of 2 kg is used for the acceleration.

Answer: (b) N > N′

Explanation:
1. Common acceleration \( a=\frac{F}{M+m+M'} \).
2. N pushes m and M′: \( N=(m+M')a \). N′ pushes only M′: \( N'=M'a \).
3. Since m > 0, N > N′.

Teacher's Note:
1. The contact force further back has to accelerate more mass.
2. Compare (m + M′)a with M′a.

Answer: (b) Statement-1 is true, Statement-2 is true, but Statement-2 is not the correct explanation of Statement-1.

Explanation:
1. On B the only horizontal force is the push of A (surface is smooth; weight and normal reaction cancel), so the force by A equals the net force on B: Statement-1 is true.
2. Statement-2 is the third law and is true, but it compares forces on two different bodies, so it does not explain why the force equals the net force on B.

Teacher's Note:
1. Third-law pairs act on different bodies and never cancel each other.
2. Net force on B means the sum of all forces acting on B.

Answer: (a) 6 m/s²

Explanation:
1. Masses: \( m_A=\frac{F}{10},\ m_B=\frac{F}{15} \).
2. Combined mass \( =F\left(\frac1{10}+\frac1{15}\right)=\frac{F}{6} \), so a = 6 m/s².

Teacher's Note:
1. Write each mass as F divided by its acceleration.
2. Add the masses and divide F by the total.

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. Total mass = 15 kg, so a = 60/15 = 4 m/s².
2. P on Q: it must accelerate Q + R + S = 8 kg, so 8 × 4 = 32 N: (i) true, (ii) false.
3. Q on R: accelerates R + S = 3 kg, so 12 N: (iii) true.
4. R pushes S with 1 × 4 = 4 N, so by Newton's third law S pushes R back with 4 N: (iv) true.

Teacher's Note:
1. The contact force at any junction equals the mass behind it times the acceleration.
2. Third-law pairs have equal magnitudes.

Answer: (b) F / sin θ

Explanation:
1. At B the three forces balance: T along AB, F horizontal, and Mg downward (through the lower rope).
2. Horizontal: T sin θ = F, so T = F / sin θ.
3. (Vertical: T cos θ = Mg, which gives the same tension as Mg / cos θ.)

Teacher's Note:
1. Resolve the tension into horizontal and vertical components.
2. Pick the direction in which only one other force acts.

Answer: (c) \( \frac{g}{7} \) downward, \( \frac{6}{7}g \)

Explanation:
1. B + A = 4 kg on one side, C = 3 kg on the other: \( a=\frac{(4-3)g}{7}=\frac{g}{7} \), with A and B moving down.
2. For A: 1·g − T = 1·(g/7), so \( T=\frac{6g}{7} \).

Teacher's Note:
1. First treat A and B as one body on one side of the pulley.
2. Use A alone to find the tension in the lower string.

Answer: (c) 240 N, 96 N

Explanation:
1. T₁ supports both bodies: T₁ = (12 + 8)(9.8 + 2.2) = 20 × 12 = 240 N.
2. T₂ supports only the 8 kg body: T₂ = 8 × (9.8 + 2.2) = 96 N.

Teacher's Note:
1. Upward acceleration adds to g in the tension: T = m(g + a).
2. Choose the system correctly: whole combination for T₁, lower body for T₂.

Answer: (c) \( \left(\frac{M_2\sin\beta-M_1\sin\alpha}{M_1+M_2}\right)g \)

Explanation:
1. Driving force along the string: M₂ g sin β down on one side, opposing M₁ g sin α on the other.
2. Net force = (M₂ sin β − M₁ sin α) g acting on total mass (M₁ + M₂).

Teacher's Note:
1. On a smooth incline the pull along the plane is mg sin(angle).
2. The same string means both blocks have the same acceleration magnitude.

Answer: (c) 2 m/s²

Explanation:
1. Net driving force = (3 − 1) g = 20 N, acting on the total mass 1 + 6 + 3 = 10 kg.
2. a = 20/10 = 2 m/s² (the 3 kg mass descends).

Teacher's Note:
1. Pulleys only change the direction of the force.
2. The table is smooth, so only the hanging weights drive the motion.

Answer: (c) Less than that in case II

Explanation:
1. Case I: \( a=\frac{2m-m}{2m+m}g=\frac{g}{3} \).
2. Case II: \( a=\frac{2mg-mg}{m}=g \).
3. So the acceleration in case I is less than in case II.

Teacher's Note:
1. In case II the rope pulls with a fixed force; in case I the force on m is smaller because the 2m mass also accelerates.
2. Compare g/3 with g.

Answer: (d) 200 N

Explanation:
1. The 2 kg and 1 kg blocks (3 kg) are pulled by the 30 N tension: a = 30/3 = 10 m/s².
2. The horizontal pull on the whole 10 kg system is F cos 60° = 10 × 10 = 100 N.
3. So F = 100 / 0.5 = 200 N.

Teacher's Note:
1. Use the tension to find the acceleration first.
2. Only the horizontal component of F accelerates the blocks.

Answer: (a) 4.9 m/s²

Explanation:
1. To lift the package the tension must be at least its weight: T = 15g.
2. For the monkey: T − 10g = 10a, so a = (15g − 10g)/10 = g/2 = 4.9 m/s².

Teacher's Note:
1. The same rope tension acts on the monkey and on the package.
2. Minimum acceleration corresponds to tension just equal to 15g.

Answer: (a) Mg

Explanation:
1. Let the left block move down with acceleration a: Mg + F − T = Ma and, for the right block, T + F − Mg = Ma.
2. Adding: 2F = 2Ma, so a = F/M.
3. Then T = Mg + F − Ma = Mg.

Teacher's Note:
1. Write one equation for each block and add them to remove T.
2. The extra forces cancel in the tension.

Answer: (d) infinitely large (not possible)

Explanation:
1. For a horizontal rope the tension has no vertical component (cos 90° = 0 in the vertical balance).
2. The weight needs 2T sin θ = 15 g, and as θ → 0 the tension must become infinite.
3. So the rope can never be made perfectly straight.

Teacher's Note:
1. Balance the vertical forces at the midpoint.
2. T = 15g / (2 sin θ) grows without limit as θ approaches zero.

Answer: (b) a₁ > a₂ > a₃

Explanation:
1. A: a₁ = (2mg − mg)/m = g.
2. B: a₂ = (2m − m)g/3m = g/3.
3. C: a₃ = (mg − mg)/m = 0.
4. So a₁ > a₂ > a₃.

Teacher's Note:
1. A force applied directly to the string can be larger than the weight carried by another mass at the same end.
2. Compute each acceleration separately.