JEE Physics Motion In A Plane MCQs Set 07

Answer: (a) \( \hat{i} \)

Explanation:
1. \( y^2 = 2x = t^2 \), so \( y = t \) as \( y > 0 \)
2. \( v_x = t,\; v_y = 1 \)
3. \( a_x = 1,\; a_y = 0 \)
4. So \( \vec a = \hat i \) at every instant. Answer: (a)

Teacher's Note:
1. Use the curve equation to express y in terms of t.
2. A velocity component that is constant has zero acceleration.

Answer: (b) 50

Explanation:
1. \( v_x = 50\cos10t \) and \( v_y = -50\sin10t \)
2. \( v = \sqrt{50^2(\cos^2 10t + \sin^2 10t)} = 50 \). Answer: (b)

Teacher's Note:
1. Differentiate each coordinate and combine the components.
2. The particle moves in a circle of radius 5 with angular speed 10, so v = r times omega.

Answer: (b) \( \cos\theta\,\hat i + \sin\theta\,\hat j \) and \( -\sin\theta\,\hat i + \cos\theta\,\hat j \)

Explanation:
1. The new x' axis makes angle \( \theta \) with the old x-axis, so \( \hat x' = \cos\theta\,\hat i + \sin\theta\,\hat j \)
2. The new y' axis makes angle \( 90^\circ + \theta \) with the old x-axis, so \( \hat y' = \cos(90^\circ+\theta)\hat i + \sin(90^\circ+\theta)\hat j = -\sin\theta\,\hat i + \cos\theta\,\hat j \)
3. The two vectors are perpendicular, as the dot product is zero. Answer: (b)

Teacher's Note:
1. Resolve each rotated unit vector along the old axes.
2. The two new unit vectors must be perpendicular.

Answer: (a) \( -8\hat i + 6\hat j \)

Explanation:
1. The magnitude of \( -4\hat i + 3\hat j \) is 5
2. Unit vector along it \( = \frac{-4\hat i + 3\hat j}{5} \)
3. Velocity \( = 10 \times \) unit vector \( = -8\hat i + 6\hat j \). Answer: (a)

Teacher's Note:
1. Velocity equals speed times the unit vector of the direction.
2. Check that the magnitude of the answer is 10.

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

Explanation:
1. \( \Delta\vec v = 5\hat j - 5\hat i \), with magnitude \( 5\sqrt2 \) m/s
2. \( a_{avg} = \frac{5\sqrt2}{10} = \frac{1}{\sqrt2} \) m/s²
3. \( \Delta\vec v \) points along \( -\hat i + \hat j \), that is towards the north-west. Answer: (b)

Teacher's Note:
1. Subtract the initial velocity vector from the final one.
2. The direction of average acceleration is that of the change in velocity.

Answer: (d) \( 120^\circ \)

Explanation:
1. \( \vec a + \vec b = \vec c \), so \(
2. \vec a + \vec b
3. ^2 = 1 \)
4. \( 1 + 1 + 2\cos\theta = 1 \), so \( \cos\theta = -\frac12 \)
5. \( \theta = 120^\circ \). Answer: (d)

Teacher's Note:
1. Square the vector equation to bring in the angle.
2. All three vectors have magnitude 1.

Answer: (c) \( \sqrt{13} \)

Explanation:
1. \(
2. \vec a+\vec b+\vec c
3. ^2 = a^2+b^2+c^2 + 2(ab+bc+ca)\cos120^\circ \)
4. \( = 4 + 9 + 36 + 2(6 + 18 + 12)(-\tfrac12) = 49 - 36 = 13 \)
5. So the magnitude is \( \sqrt{13} \). Answer: (c)

Teacher's Note:
1. Expand the square of the sum using dot products.
2. cos 120 degrees equals minus one half.

Answer: (d) \( 3t^2\sqrt{\alpha^2+\beta^2} \)

Explanation:
1. \( v_x = 3\alpha t^2 \) and \( v_y = 3\beta t^2 \)
2. \( v = \sqrt{v_x^2 + v_y^2} = 3t^2\sqrt{\alpha^2+\beta^2} \). Answer: (d)

Teacher's Note:
1. Speed is the magnitude of the velocity vector.
2. Differentiate each coordinate separately.

Answer: (b) k/3

Explanation:
1. The horizontal component stays \( v\cos60^\circ = \frac v2 \)
2. At 30°, \( v'\cos30^\circ = \frac v2 \), so \( v' = \frac{v}{\sqrt3} \)
3. \( K' = \frac12 m \frac{v^2}{3} = \frac k3 \). Answer: (b)

Teacher's Note:
1. The horizontal component of velocity is constant in projectile motion.
2. Kinetic energy is proportional to the square of the speed.

Answer: (d) 25 m

Explanation:
1. Compare \( y = u_y t - \tfrac12 g t^2 \) with \( y = 10t - t^2 \): \( u_y = 10 \) m/s and \( g = 2 \) m/s²
2. \( H = \frac{u_y^2}{2g} = \frac{100}{4} = 25 \) m. Answer: (d)

Teacher's Note:
1. Read the vertical velocity and the planet's g directly from the equation.
2. The planet's g is not 9.8, so do not substitute Earth's value.