JEE Physics Kinematics MCQs Set 05

Answer: (b) 1, 2, 3 and 4 are all correct

Explanation:
1. Constant acceleration only along y means \( v_x \) is constant. (1) is correct.
2. Comparing with \( y = x\tan\theta - \frac{gx^2}{2v_x^2} \): \( a = \tan\theta \) and \( b = \frac{g}{2v_x^2} \), so \( v_x = \sqrt{\frac{g}{2b}} \) and \( v_y = av_x = a\sqrt{\frac{g}{2b}} \). (2) and (3) are correct.
3. A constant downward acceleration is exactly what defines projectile motion. (4) is correct.

Teacher's Note:
1. Match the trajectory with the standard projectile equation.
2. tan θ is the initial slope of the path.

Answer: (c) 1, 3 and 4 only

Explanation:
1. (1) \( h = \tfrac12(10)(0.4)^2 = 0.8 \) m. (3) \( x = 4\times0.4 = 1.6 \) m. (4) \( v_y = 10\times0.4 = 4 \) m/s.
2. (2) is wrong: \( \tan\phi = \frac{v_x}{v_y} = 1 \), so the angle with the vertical is 45°.

Teacher's Note:
1. Horizontal and vertical velocities are equal at the ground, so the path makes 45°.
2. Time of fall depends only on the height.

Answer: (d) 1, 2 and 3 only

Explanation:
1. The displacement is zero, so the average velocity is zero. (1) and (2) are true.
2. Total distance \( = 2\times\frac{u^2}{2g} = \frac{u^2}{g} \) in time \( \frac{2u}{g} \), so the average speed is u/2. (3) is true and (4) is false.

Teacher's Note:
1. Average speed uses the total path length.
2. Speed varies from u to 0 and back to u, with mean u/2.

Answer: (b) 2, 3 and 4 only

Explanation:
1. The relative speed 6 m/s falls to zero in 4 s, so the acceleration is \( 1.5 \) m/s² towards the belt direction. Velocity w.r.t. ground: \( v = 2 - 1.5t \).
2. \( x = 2t - 0.75t^2 \). It is zero at \( t = \frac83 \) s (4 is correct) and equals \( 8 - 12 = -4 \) m at t = 4 s. So (2) is correct and the 12 m claim in (1) is wrong.
3. With respect to the belt: \( \tfrac12(6)(4) = 12 \) m. (3) is correct.

Teacher's Note:
1. The displacement relative to the belt is the area of the relative velocity graph.
2. Relative to the ground the block turns back.

Answer: (a) 3 and 4 only

Explanation:
1. With constant velocity the ball shares the cart's horizontal velocity and returns to the thrower.
2. If the cart accelerates, the ball keeps its horizontal speed while the cart speeds up, so the ball falls behind. So (3) and (4) are true, (1) and (2) are false.

Teacher's Note:
1. The horizontal velocity of the ball stays what it was at release.
2. The result depends on whether the cart accelerates.

Answer: (c) 2 and 4 only

Explanation:
1. A swims straight across (width 10 m in 120 s), so its speed relative to the river is 10/120 = 1/12 m/s. (1) is wrong. The river carries A 30 m downstream in 120 s, so the flow speed is 30/120 = 1/4 m/s. (2) is correct.
2. B also covers 10 m across in 120 s, so its across-river ground velocity is 1/12 m/s, not 1/3 m/s. (3) is wrong. B covers 25 m along the bank in 120 s, so its ground velocity along the flow is 25/120 = 5/24 m/s. (4) is correct.

Teacher's Note:
1. The ground velocity components follow from displacement divided by time.
2. Both swimmers cross the same width in the same time.

Answer: (a) 1 and 3 only

Explanation:
1. vx = 40 cos θ = 32 m/s and initial vy = 24 m/s. After 2 s, vy = 24 − 20 = 4 m/s.
2. tan α = 4/32 = 1/8 and \( v = \sqrt{32^2+4^2} = \sqrt{1040} \approx 32.25 \) m/s. So (1) and (3) are correct.

Teacher's Note:
1. tan θ = 3/4 gives sin θ = 3/5 and cos θ = 4/5.
2. The speed is not equal to vx because vy is not zero.

Answer: (d) 3 and 4 only

Explanation:
1. The same maximum height means the same vertical velocity component, so the times of flight \( T = \frac{2u_y}{g} \) are equal: \( T_1 = T_2 = T_3 \). (4) is correct.
2. Since the three times are equal, T₂ = (T₁ + T₃)/2 also holds. (3) is correct. (1) and (2) are wrong.

Teacher's Note:
1. Time of flight depends only on the vertical component of the launch velocity.
2. The different ranges come from different horizontal velocities.

Answer: (b) \( \frac{1}{\sqrt2} \) m/s² towards the north-west

Explanation:
1. \( \Delta\vec v = 5\hat j - 5\hat i \), with magnitude \( 5\sqrt2 \) m/s and directed towards the north-west.
2. \( \vec a_{avg} = \frac{\Delta\vec v}{10} = \frac{1}{\sqrt2} \) m/s² towards the north-west.

Teacher's Note:
1. Subtract the vectors, do not subtract the speeds.
2. The change in velocity points from the tip of the initial vector to the tip of the final vector.

Answer: (c) \( s = \frac{1}{72}ft^2 \)

Explanation:
1. Speed after the first stage: \( v^2 = 2fs \). Distance while decelerating at f/2: \( \frac{v^2}{2(f/2)} = \frac{v^2}{f} = 2s \).
2. Total: \( s + vt + 2s = 15s \), so \( vt = 12s \) and \( 2fs\,t^2 = 144s^2 \).
3. Hence \( s = \frac{ft^2}{72} \).

Teacher's Note:
1. Write the distance of each of the three stages.
2. Square vt = 12s to eliminate v.

Answer: (a) \( -2av^3 \)

Explanation:
1. \( \frac{dt}{dx} = 2ax + b = \frac1v \), so \( v = \frac{1}{2ax+b} \).
2. \( \frac{dv}{dt} = \frac{dv}{dx}\cdot v = -\frac{2a}{(2ax+b)^2}\cdot v = -2av^3 \).

Teacher's Note:
1. Differentiate t with respect to x first, which gives 1/v.
2. Then use a = v dv/dx.

Answer: (d) R

Explanation:
1. The angles are θ and 90° − θ: \( t_1 = \frac{2u\sin\theta}{g} \) and \( t_2 = \frac{2u\cos\theta}{g} \).
2. \( t_1t_2 = \frac{4u^2\sin\theta\cos\theta}{g^2} = \frac{2R}{g} \), so it is proportional to R.

Teacher's Note:
1. Complementary angles give equal ranges.
2. 2 sinθ cosθ = sin 2θ.

Answer: (b) \( t^2 \)

Explanation:
1. \( \frac{dx}{dt} = \alpha\sqrt x \Rightarrow \int x^{-1/2}dx = \int\alpha\,dt \Rightarrow 2\sqrt x = \alpha t \).
2. So \( x = \frac{\alpha^2t^2}{4} \propto t^2 \).

Teacher's Note:
1. Separate the variables and integrate.
2. Constant acceleration α²/2 gives the same result.

Answer: (d) K/4

Explanation:
1. At the top only the horizontal component \( u\cos60^\circ = \frac u2 \) remains.
2. So \( KE = \frac12m\left(\frac u2\right)^2 = \frac K4 \).

Teacher's Note:
1. Kinetic energy depends on the square of the speed.
2. cos²60° = 1/4.

Answer: (a) \( v_0 + \frac g2 + \frac f3 \)

Explanation:
1. \( x = \int_0^1(v_0 + gt + ft^2)dt = v_0 + \frac g2 + \frac f3 \).

Teacher's Note:
1. Integrate each term with the correct power.
2. Do not substitute t = 1 directly into the velocity.

Answer: (c) \( 7\sqrt2 \) units

Explanation:
1. \( \vec v = (3 + 0.4\times10)\hat i + (4 + 0.3\times10)\hat j = 7\hat i + 7\hat j \).
2. Speed \( = 7\sqrt2 \).

Teacher's Note:
1. Add the vector change in velocity to the initial velocity.
2. Take the magnitude only at the end.

Answer: (c) \( y^2 = x^2 + \text{constant} \)

Explanation:
1. \( \frac{dx}{dt} = ky \) and \( \frac{dy}{dt} = kx \), so \( \frac{dy}{dx} = \frac xy \).
2. \( y\,dy = x\,dx \Rightarrow y^2 = x^2 + c \).

Teacher's Note:
1. The path is found from dy/dx = vy/vx.
2. The curve is a hyperbola.

Answer: (a) \( \pi\frac{v^4}{g^2} \)

Explanation:
1. The farthest point is reached at 45°: \( R_{max} = \frac{v^2}{g} \).
2. The wet area is a circle of radius R_max: \( \pi R_{max}^2 = \pi\frac{v^4}{g^2} \).

Teacher's Note:
1. Maximum range occurs at 45°.
2. Area = π (radius)².

Answer: (d) 2 s

Explanation:
1. \( \int_{6.25}^0\frac{dv}{\sqrt v} = -2.5\int_0^tdt \).
2. \( 2\left[\sqrt v\right]_{6.25}^{0} = -5 = -2.5t \), so t = 2 s.

Teacher's Note:
1. Separate the variables v and t.
2. √6.25 = 2.5.

Answer: (b) \( \frac{\sqrt3}{16}\frac{mv^3}{g} \)

Explanation:
1. At the top the velocity is horizontal, \( v\cos30^\circ = \frac{\sqrt3v}{2} \), and the height is \( h = \frac{v^2\sin^230^\circ}{2g} = \frac{v^2}{8g} \).
2. \( L = m(v\cos30^\circ)h = m\cdot\frac{\sqrt3v}{2}\cdot\frac{v^2}{8g} = \frac{\sqrt3mv^3}{16g} \).

Teacher's Note:
1. Angular momentum = m × (velocity component perpendicular to r) × r, or m v_horizontal × h here.
2. The perpendicular distance of the horizontal velocity from the origin is h.

Answer: (b) 20 m

Explanation:
1. The speed of throw satisfies \( \frac{u^2}{2g} = 10 \), so \( \frac{u^2}{g} = 20 \) m.
2. The maximum range is u²/g = 20 m.

Teacher's Note:
1. The same throwing speed is used.
2. The maximum range is twice the maximum height.

Answer: (c) \( \frac{F_0}{mb}\left(1 - e^{-bt}\right) \)

Explanation:
1. \( v = \frac1m\int_0^tF_0e^{-bt}dt = \frac{F_0}{mb}\left(1 - e^{-bt}\right) \).
2. The speed rises from zero with decreasing slope and approaches \( \frac{F_0}{mb} \).

Teacher's Note:
1. Integrate the acceleration F/m.
2. Check t = 0 (v = 0) and t → ∞ (v → F₀/mb).

Answer: (a) \( y = 2x - 5x^2 \)

Explanation:
1. \( x = t \) and \( y = 2t - 5t^2 \).
2. Eliminating t: \( y = 2x - 5x^2 \).

Teacher's Note:
1. Eliminate time between the two coordinate equations.
2. The coefficient 5 is g/2.

Answer: (d) \( 2gH = nu^2(n-2) \)

Explanation:
1. The time to the highest point is \( \frac ug \), so the time to hit the ground is \( \frac{nu}{g} \).
2. Taking up as positive: \( -H = u\cdot\frac{nu}{g} - \frac12g\frac{n^2u^2}{g^2} \), so \( H = \frac{u^2}{g}\left(\frac{n^2}{2} - n\right) \).
3. Hence \( 2gH = nu^2(n-2) \).

Teacher's Note:
1. Displacement is −H at the instant the particle reaches the ground.
2. For n = 2 the particle returns to the tower top and H = 0, as expected.

Answer: (b) negative at x = 0 and increasing linearly to zero at x = x₁

Explanation:
1. \( v = v_0\left(1 - \frac{x}{x_1}\right) \).
2. \( a = v\frac{dv}{dx} = -\frac{v_0^2}{x_1}\left(1 - \frac{x}{x_1}\right) \).
3. This is negative at x = 0 and rises linearly to zero at x = x₁.

Teacher's Note:
1. Use a = v dv/dx.
2. The slope of v-x is constant, but v itself falls, so a falls in magnitude.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Statement (R) is the correct formula for relative velocity, so it is true.
2. Statement (A) is also true as a real observation. But the relative velocity formula alone does not tell why distant objects seem still: that depends on the angle subtended at the eye, which changes slowly for far objects and quickly for near ones. So R does not explain A.

Teacher's Note:
1. Relative velocity gives the actual relative speed, which is the same for near and far objects.
2. The apparent motion depends on angular speed, which falls with distance.

Answer: (d) 5

Explanation:
1. Time of flight \( T = \frac{2(10)\sin60^\circ}{10} = \sqrt3 \) s.
2. In the train frame the ball has horizontal velocity \( 10\cos60^\circ = 5 \) m/s and acceleration −a. \( 5T - \tfrac12aT^2 = 1.15 \Rightarrow 5\sqrt3 - 1.5a = 1.15 \).
3. \( a = \frac{8.66 - 1.15}{1.5} \approx 5 \) m/s².

Teacher's Note:
1. The ball returns to the boy when its horizontal position relative to the train equals his displacement.
2. In the train's frame the ball feels a backward pseudo acceleration a.