JEE (Joint Entrance Examination):

1. A 50-kg mass is acted upon by a force of 100 N. What is the acceleration of the mass? Shortcut: Use Newton’s second law (F = ma) Acceleration (a) = Force (F) / Mass (m) Substituting the given values: a = 100 N / 50 kg = 2 m/s² Therefore, the acceleration of the mass is 2 m/s².

2. A 2-kg mass is moving at a velocity of 10 m/s. What is the force required to bring the mass to rest in 5 seconds? Shortcut: Use the equation: Force (F) = Mass (m) × (Change in Velocity) / Time First, calculate the change in velocity: (Final velocity - Initial velocity). Since the mass needs to be brought to rest, the final velocity will be 0 m/s. Change in Velocity = 0 m/s - 10 m/s = -10 m/s Now substitute the values in the equation: F = 2 kg × (-10 m/s) / 5 s = -4 N The negative sign indicates that the force should act in the opposite direction to the motion to bring the mass to rest. Therefore, a force of 4 N acting in the opposite direction is required to bring the mass to rest in 5 seconds.

3. A 10-kg mass is suspended from a spring. The spring constant is 100 N/m. What is the frequency of oscillation of the mass? Shortcut: Use the formula: Frequency (f) = 1 / (2π) × √(Spring Constant (k) / Mass (m)) Substitute the given values: f = 1 / (2π) × √(100 N/m / 10 kg) f ≈ 0.159 Hz Therefore, the frequency of oscillation of the mass is approximately 0.159 Hz.

CBSE (Central Board of Secondary Education):

1. A 100-g ball is thrown vertically upward with a speed of 20 m/s. How high will the ball go? Shortcut: Use the equation: Maximum Height (h) = (Initial Velocity)² / (2 × Acceleration due to gravity) Substitute the given values: h = (20 m/s)² / (2 × 9.8 m/s²) = 20.41 m Therefore, the ball will reach a maximum height of approximately 20.41 meters.

2. A 5-kg block is pulled across a horizontal surface by a force of 20 N. The coefficient of kinetic friction is 0.2. What is the acceleration of the block? Shortcut: Use Newton’s second law (F = ma) along with the formula for frictional force (F_friction = Coefficient of Kinetic Friction × Normal Force) First, calculate the normal force acting on the block. Since the block is on a horizontal surface, the normal force is equal to the weight of the block: Normal Force = Mass × Acceleration due to gravity Normal Force = 5 kg × 9.8 m/s² = 49 N Next, calculate the frictional force: F_friction = Coefficient of Kinetic Friction × Normal Force F_friction = 0.2 × 49 N = 9.8 N Now, substitute the values in Newton’s second law, considering the frictional force as the opposing force: 20 N - 9.8 N = 5 kg × Acceleration 10.2 N = 5 kg × Acceleration Acceleration = 2.04 m/s² Therefore, the acceleration of the block is approximately 2.04 m/s².

3. A 2-kg mass is attached to a spring. The spring constant is 50 N/m. What is the period of oscillation of the mass? Shortcut: Use the formula: Time Period (T) = 2π × √(Mass (m) / Spring Constant (k)) Substitute the given values: T = 2π × √(2 kg / 50 N/m) T ≈ 0.63