JEE Physics Laws Of Motion MCQs Set 12

Answer: (c) \( -\vec{F}/m \)

Explanation:
1. Before cutting, the net force is zero, so the other forces together equal \( -\vec{F} \).
2. Spring forces do not change instantly, so just after the cut the net force is \( -\vec{F} \).
3. The acceleration is \( -\vec{F}/m \).

Teacher's Note:
1. A spring force does not change at the instant of cutting another spring.
2. The remaining forces add up to minus F.

Answer: (c) 3000 N/m

Explanation:
1. Let the spring force be \( T_s \) and the string tension \( T \). Horizontal balance: \( T_s\cos37° = T\cos53° \), so \( T = \tfrac43 T_s \).
2. Vertical balance: \( T_s\sin37° + T\sin53° = 200 \), giving \( 0.6T_s + 0.8 \times \tfrac43 T_s = 200 \), so \( T_s = 120 \) N.
3. \( k = 120/0.04 = 3000 \) N/m.

Teacher's Note:
1. Use equilibrium of the knot in both directions.
2. Convert 4 cm into 0.04 m.

Answer: (b) \( T_1 : T_2 : T_3 :: 6 : 5 : 3 \)

Explanation:
1. 0 to 2 s: \( a = +2 \), so \( T_1 = 100(10 + 2) = 1200 \) N.
2. 2 to 6 s: \( a = 0 \), so \( T_2 = 1000 \) N.
3. 6 to 7 s: \( a = -4 \), so \( T_3 = 100(10 - 4) = 600 \) N.
4. Ratio \( = 1200 : 1000 : 600 = 6 : 5 : 3 \).

Teacher's Note:
1. Find the acceleration from the slope of the speed-time graph in each interval.
2. Tension is m(g + a).

Answer: (b) less than 3 kg

Explanation:
1. The acceleration of the masses is \( a = g/3 \).
2. Tension \( T = \dfrac{2 \times 1 \times 2}{1 + 2}g = \dfrac{4g}{3} \).
3. The balance reads \( 2T = \dfrac{8}{3} \) kg-wt \( = 2.67 \) kg, which is less than 3 kg.

Teacher's Note:
1. The balance reads the sum of the two tensions on the pulley.
2. Part of the weight goes into accelerating the masses.

Answer: (b) \( x_2 > x_1 > x_3 \)

Explanation:
1. For an Atwood arrangement, the tension is \( T = \dfrac{2m_1m_2}{m_1 + m_2}g \), and the spring carries this tension.
2. Case 1: \( T = 2g \). Case 2: \( T = \tfrac{12}{5}g = 2.4g \). Case 3: \( T = \tfrac43 g \approx 1.33g \).
3. Extension is proportional to tension, so \( x_2 > x_1 > x_3 \).

Teacher's Note:
1. The spring extension depends on the tension.
2. The tension is largest when the masses are both large and nearly unequal.

Answer: (b) \( \sqrt{40} \) N

Explanation:
1. Net force \( = (3 - 1)\hat{i} + (-4 - 2)\hat{j} = 2\hat{i} - 6\hat{j} \) N.
2. Magnitude \( = \sqrt{2^2 + 6^2} = \sqrt{40} \) N.

Teacher's Note:
1. Add the components first, then take the magnitude.
2. Do not add the magnitudes.

Answer: (c) 2 N, 8 N

Explanation:
1. First case: \( a = F/9m \) and \( P = (3m + 5m)a = 8ma = 16 \) N, so \( ma = 2 \) N.
2. Second case: the acceleration is again \( F/9m \), so \( ma = 2 \) N still.
3. \( P' = ma = 2 \) N (pulls only m) and \( Q' = (m + 3m)a = 4 \times 2 = 8 \) N.

Teacher's Note:
1. The common acceleration is the same in both cases.
2. Tension equals the mass behind times the acceleration.

Answer: (b) g/2

Explanation:
1. In the lift frame the effective gravity is \( g + a \).
2. Relative acceleration \( = \dfrac{2m - m}{2m + m}(g + a) = \dfrac{g + a}{3} \).
3. Setting this equal to \( g/2 \): \( g + a = 3g/2 \), so \( a = g/2 \).

Teacher's Note:
1. In an accelerating lift replace g by g plus a.
2. The relative acceleration is the Atwood result with the effective gravity.

Answer: (a) \( (m_1 + m_2 + m_3)\dfrac{m_2 g}{m_1} \)

Explanation:
1. \( m_2 \) does not move vertically, so the string tension \( T = m_2 g \).
2. This tension is the only horizontal force on \( m_1 \), so \( m_1 a = m_2 g \) and \( a = m_2 g/m_1 \).
3. All three move together, so \( F = (m_1 + m_2 + m_3)a = (m_1 + m_2 + m_3)\dfrac{m_2 g}{m_1} \).

Teacher's Note:
1. The hanging block presses on the cart side and gets its horizontal acceleration from the cart.
2. Use one common acceleration for the whole system.

Answer: (c) 2M/5

Explanation:
1. Let the buoyant force be B. Descending: \( Mg - B = M\,g/4 \), so \( B = 3Mg/4 \).
2. After releasing m: \( B - (M - m)g = (M - m)g/4 \), so \( B = \tfrac54 (M - m)g \).
3. \( \tfrac34 M = \tfrac54 (M - m) \), so \( M - m = 3M/5 \) and \( m = 2M/5 \).

Teacher's Note:
1. The buoyant force does not change when mass is removed.
2. Write Newton's second law before and after.

Answer: (d) \( \dfrac{v}{\cos\theta} \)

Explanation:
1. The rope is inextensible, so the component of the block's velocity along the rope equals v.
2. The block moves horizontally with speed u: \( u\cos\theta = v \).
3. So \( u = v/\cos\theta \).

Teacher's Note:
1. Constraint relations come from the fixed length of the string.
2. The velocity component along the string is the same at both ends.

Answer: (c) \( \dfrac{\cos\theta_2}{\cos\theta_1} \)

Explanation:
1. The string length is constant, so the rate at which the string leaves one block equals the rate at which it goes to the other.
2. The components along the strings are \( v_1\cos\theta_1 \) and \( v_2\cos\theta_2 \), so they are equal.
3. \( v_1/v_2 = \cos\theta_2/\cos\theta_1 \).

Teacher's Note:
1. Equal lengths of string pass the pulley on each side.
2. Use the components along the string.

Answer: (b) \( v/\cos\theta \)

Explanation:
1. The string length from each pulley to the mass shortens at rate v, because the end P (or Q) goes down at v.
2. The mass moves up with speed u, and the string is inclined at θ to the vertical, so the shortening rate is \( u\cos\theta \).
3. So \( u\cos\theta = v \) and \( u = v/\cos\theta \).

Teacher's Note:
1. Resolve the velocity of the mass along the string.
2. The rate of shortening of each inclined part equals v.

Answer: (a) g downward, g upward

Explanation:
1. Before cutting, the spring force is \( 2mg \) and tension in the string is \( mg \).
2. After cutting, block A is only under gravity, so its acceleration is g downward.
3. The spring force cannot change at once, so B has net force \( 2mg - mg = mg \) upward, giving g upward.

Teacher's Note:
1. A string tension disappears at once while a spring force does not.
2. Treat the two blocks separately after the cut.

Answer: (b) 6.92 N

Explanation:
1. The horizontal string tension equals the friction, at most 12 N.
2. At the knot: \( T\cos30° = 12 \) and \( T\sin30° = W \).
3. \( W = 12\tan30° = 12/\sqrt{3} = 6.92 \) N.

Teacher's Note:
1. Resolve the forces at the knot.
2. Tension in the horizontal string is limited by the friction.

Answer: (b) 11 kg

Explanation:
1. Uniform speed means zero net force on each block.
2. The tension equals \( M_2 g \), and it balances friction: \( M_2 g = \mu(M_1 + m)g \).
3. \( 6 = 0.4(4 + m) \), so \( 4 + m = 15 \) and \( m = 11 \) kg.

Teacher's Note:
1. For uniform speed, driving force equals friction.
2. Extra load on the block increases the normal reaction and so the friction.

Answer: (b) 42 N

Explanation:
1. Maximum friction \( = 0.5 \times 7 \times 10 = 35 \) N.
2. At the limit the slab has \( a = 35/35 = 1 \) m/s², and the block has the same acceleration.
3. \( F - 35 = 7 \times 1 \), so \( F = 42 \) N.

Teacher's Note:
1. At the limit both bodies move together with the maximum friction acting.
2. The slab is accelerated only by friction.

Answer: (a) zero

Explanation:
1. Limiting friction \( = 0.5 \times 40 \times 10 = 200 \) N.
2. If no slipping occurs, \( a = 40/50 = 0.8 \) m/s² and the friction needed on the slab is \( 10 \times 0.8 = 8 \) N, which is far below 200 N.
3. So there is no relative motion, and the relative acceleration is zero.

Teacher's Note:
1. Check that required friction is below the limiting value.
2. If it is, both bodies move together.

Answer: (b) 50 N

Explanation:
1. The graph is a straight line from 0 at t = 0 to 4 m/s at t = 2 s.
2. Acceleration \( a = 4/2 = 2 \) m/s².
3. \( F = ma = 25 \times 2 = 50 \) N.

Teacher's Note:
1. The slope of a velocity-time graph is the acceleration.
2. Then apply F = ma.

Answer: (a) zero

Explanation:
1. Area from 0 to 4 s: triangle \( = \tfrac12 \times 4 \times 3 = 6 \) N s.
2. Area from 4 s to 8 s: triangle 4 to 5 s \( = -1 \), rectangle 5 to 7 s \( = -4 \), triangle 7 to 8 s \( = -1 \), a total of \( -6 \) N s.
3. Net change in momentum \( = 6 - 6 = 0 \).

Teacher's Note:
1. Impulse equals the area under the F-t graph.
2. Positive and negative areas cancel here.

Answer: (c) 14.3 m/s

Explanation:
1. Area under the graph: 0 to 2 s \( = \tfrac12 \times 2 \times 4 = 4 \). 2 to 4 s \( = 4 \times 2 = 8 \).
2. 4 to 4.5 s \( = \tfrac12(4 + 2.5)\times0.5 = 1.625 \). 4.5 to 6.5 s \( = 2.5 \times 2 = 5 \). Total impulse \( = 18.625 \) N s.
3. \( \Delta v = 18.625/2 = 9.31 \) m/s, so the final speed \( = 5 + 9.31 \approx 14.3 \) m/s.

Teacher's Note:
1. Impulse is the area under the force-time graph.
2. Add the change in velocity to the initial speed.

Answer: (b) \( -0.2 \) kg m/s

Explanation:
1. For 0 to 2 s the graph is a straight line from 0 to 4 m, so the velocity is 2 m/s.
2. After 2 s the position is constant, so the velocity is zero.
3. Impulse \( = \Delta p = 0.1(0) - 0.1(2) = -0.2 \) kg m/s.

Teacher's Note:
1. The slope of a position-time graph is velocity.
2. Impulse is the change in momentum.

Answer: (d) 5 N

Explanation:
1. Both blocks have mass 1 kg and are joined by the same string, so they move with the same acceleration.
2. The hanging block pulls with its weight 10 N on a total mass of 2 kg, so \( a = 10/2 = 5 \) m/s².
3. For the hanging block \( 10 - T = 1 \times 5 \), so \( T = 5 \) N.

Teacher's Note:
1. The string joins the two blocks, so the acceleration is common.
2. Write Newton's law for the hanging block to get the tension.

Answer: (c) \( \dfrac{2g\sin\theta}{3} \)

Explanation:
1. The driving force is the component of the weight of 2M along its plane: \( 2Mg\sin\theta \).
2. The string and the right face carry no resistance because all surfaces are smooth.
3. \( a = \dfrac{2Mg\sin\theta}{2M + M} = \dfrac{2g\sin\theta}{3} \).

Teacher's Note:
1. Only the weight component along the incline drives the system.
2. The total mass moving is 3M.