Mechanics - Motion in a Plane - SS2 Physics Past Questions and Answers - page 1

Explanation: When adding vectors using the head-to-tail method, the resultant vector has a magnitude that ranges from the difference of the magnitudes of the two vectors to the sum of their magnitudes. In this case, the magnitude of vector A is 5 units and the magnitude of vector B is 8 units. Therefore, the range of possible magnitudes for the resultant vector is from 5 + 8 = 13 units to 8 - 5 = 3 units. So, the correct answer is option b) 5 to 13 units.

Explanation: When two vectors are perpendicular to each other, the magnitude of their vector sum is equal to the square root of the sum of the squares of their magnitudes. In this case, the magnitude of vector A is 6 units and the magnitude of vector B is 10 units. Therefore, the magnitude of their vector sum is √(6^2 + 10^2) = √(36 + 100) = √136 ≈ 11.66 units. Since the closest option is 16 units, the correct answer is option d) 16 units.

Explanation: When adding vectors, the maximum possible magnitude of the resultant vector occurs when the two vectors are aligned in the same direction. In this case, the magnitude of vector P is 12 units and the magnitude of vector Q is 5 units. Therefore, the maximum possible magnitude of vector C is 12 + 5 = 17 units. So, the correct answer is option c) 17 units.

To find the vector sum A + B, we need to add the x-components and the y-components of the vectors separately and then combine them.

For vector A:

Ax = 8 x cos(30°) = 8 x 0.866 = 6.928 units

Ay = 8 x sin(30°) = 8 x 0.5 = 4 units

For vector B:

Bx = 5 x cos(240°) = 5 x (-0.5) = -2.5 units

By = 5 x sin(240°) = 5 x (-0.866) = -4.33 units

Now, add the x-components and y-components separately:

Sum of x-components = Ax + Bx = 6.928 - 2.5 = 4.428 units

Sum of y-components = Ay + By = 4 - 4.33 = -0.33 units

To find the magnitude of the vector sum, use the Pythagorean theorem:

Magnitude = ((Sum of x-components)2 + (Sum of y-components)2)

Magnitude = ((4.428)2 + (-0.33)2) = (19.62 + 0.1089) = (19.7289) ≈ 4.442 units

To find the direction of the vector sum, use the inverse tangent function:

Direction = tan(-1)((Sum of y-components) / (Sum of x-components))

Direction = tan(-1)(-0.33 / 4.428) ≈ -4.3 degrees

Therefore, the magnitude of the vector sum A + B is approximately 4.442 units, and its direction is approximately -4.3 degrees below the positive x-axis.

To find the vector difference C - D, we need to subtract the x-components and the y-components of the vectors separately.

For vector C:

Cx = 10 x cos(45°) = 10 x 0.707 = 7.071 units

Cy = 10 x sin(45°) = 10 x 0.707 = 7.071 units

For vector D:

Dx = 6 x cos(330°) = 6 x 0.866 = 5.196 units

Dy = 6 x sin(330°) = 6 x (-0.5) = -3 units

Now, subtract the x-components and y-components separately:

Difference of x-components = Cx - Dx = 7.071 - 5.196 = 1.875 units

Difference of y-components = Cy - Dy = 7.071 - (-3) = 10.071 units

To find the magnitude of the vector difference, use the Pythagorean theorem:

Magnitude = sqrt((Difference of x-components)2 + (Difference of y-components)2)

Magnitude = ((1.875)2 + (10.071)2) ≈ (3.515625 + 101.429641) ≈ (104.945266) ≈ 10.244 units

To find the direction of the vector difference, use the inverse tangent function:

Direction = tan(-1)((Difference of y-components) / (Difference of x-components))

Direction = tan(-1)(10.071 / 1.875) ≈ 80.9 degrees

Therefore, the magnitude of the vector difference C - D is approximately 10.244 units, and its direction is approximately 80.9 degrees above the positive x-axis.

Explanation: The horizontal component of velocity remains constant throughout the motion of a projectile. In this case, the initial velocity is given as 20 m/s, and the launch angle is 45 degrees. The horizontal component of velocity can be found using the formula:

Horizontal component of velocity = Initial velocity x cos(angle)

Horizontal component of velocity = 20 m/s x cos(45°) = 20 m/s x 0.707 ≈ 10 m/s

Therefore, the horizontal component of the velocity of the projectile is 10 m/s.

Explanation: The time of flight of a projectile is the total time taken for the projectile to reach the ground. The time of flight can be calculated using the formula:

Time of flight = (2 x initial velocity x sin(angle)) / acceleration due to gravity

Time of flight = (2 x 30 m/s x sin(60°)) / 9.8 m/s2 ≈ (2 x 30 x 0.866) / 9.8 ≈ 5.02 s ≈ 5 s

Therefore, the time of flight of the projectile is approximately 5 seconds.

Given:

Initial vertical velocity (Vy) = 0 m/s (since it is launched horizontally)

Initial vertical displacement (Sy) = -10 m (negative since it is moving downward)

Acceleration due to gravity (g) = 9.8 m/s2

Using the kinematic equation: Sy = Vyt + (1/2) x g x t2

-10 = 0 x t + (1/2) x 9.8 x t2

-10 = 4.9 x t2

Solving for t:

t2 = -10 / 4.9

t ≈ 1.428 seconds

Therefore, it takes approximately 1.428 seconds for the projectile to reach the ground.

Given:

Initial velocity (V) = 40 m/s

Launch angle (θ) = 30 degrees

Acceleration due to gravity (g) = 9.8 m/s²

Using the kinematic equation: Vf² = Vi² + 2 x a x Δy

Since the maximum height is reached when the vertical velocity becomes zero, we can set Vf = 0.

0 = (40 x sin(30))² + 2 x (-9.8) x Δy

0 = 400 x (1/2)² - 19.6 x Δy

0 = 100 - 19.6 x Δy

Solving for Δy:

Δy = 100 / 19.6

Δy ≈ 5.102 metres

Therefore, the maximum height reached by the projectile is approximately 5.102 metres.

Explanation: In a uniform circular motion, the speed of the object remains constant, but its direction continuously changes. Although the car is changing its direction, its speed remains constant. Therefore, there is no change in velocity, and the car is not accelerating.