Applied Mathematics — Second: Dynamics · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 37 questions on this lesson, each with an explanation.
A body of mass 5 kg starts from rest and reaches a speed of 6 m/sec. Find the impulse given to it.
Correct answer: (D) 30 N.sec
Explanation: Step 1: Impulse I = m(v − u), with u = 0. Step 2: I = 5 × (6 − 0) = 30. ∴ Impulse = 30 N.sec
A constant force of 15 N acts on a body for 4 seconds. Find the impulse of the force.
Correct answer: (C) 60 N.sec
Explanation: Step 1: Impulse of a constant force I = F t. Step 2: I = 15 × 4 = 60. ∴ Impulse = 60 N.sec
A ball of mass 150 gm moving horizontally at 12 m/sec hits a smooth vertical wall and rebounds along the same line at 8 m/sec. The time of contact is 0.05 sec. Find the pressure of the ball on the wall.
Correct answer: (D) 60 N
Explanation: Taking the rebound direction as positive: impulse I = m(v₂ − v₁) = 0.15 × (8 − (−12)) = 3 N.sec. I = F t ⇒ F = 3 ÷ 0.05 = 60 N, and for a horizontal impact the pressure on the wall equals F. (12 N comes from using 12 − 8 instead of 12 + 8; 3 N is the impulse, not the force.)
A machine gun fires 300 bullets per minute horizontally, each of mass 40 gm, with a velocity of 720 km/h. Find the magnitude of the average force of reaction on the gun.
Correct answer: (C) 40 N
Explanation: Step 1: v = 720 × 5/18 = 200 m/sec and m = 0.04 kg, so the momentum given to one bullet = 0.04 × 200 = 8 N.sec. Step 2: In one second the gun fires 300 ÷ 60 = 5 bullets, so the impulse per second = 5 × 8 = 40 N.sec. Step 3: Force = impulse ÷ time = 40 ÷ 1 = 40 N (the reaction on the gun is equal and opposite).
A body of mass 4 kg has its velocity changed from (2 î + 3 ĵ) m/sec to (5 î + 7 ĵ) m/sec. Find the magnitude of the impulse on it.
Correct answer: (C) 20 N.sec
Explanation: Step 1: Change in velocity: (5 − 2) î + (7 − 3) ĵ = 3 î + 4 ĵ, with magnitude √(3² + 4²) = 5 m/sec. Step 2: Impulse I = m × (change in velocity), so ‖I‖ = 4 × 5. Step 3: ‖I‖ = 20. ∴ Impulse = 20 N.sec
A body of mass 2 kg moving at 5 m/sec strikes a barrier and rebounds at 5 m/sec in the opposite direction. Find the magnitude of the impulse on the body.
Correct answer: (D) 20 N.sec
Explanation: Step 1: Take the rebound direction as positive: u = −5 m/sec, v = +5 m/sec. Step 2: I = m(v − u) = 2 × (5 − (−5)) = 2 × 10. Step 3: I = 20. ∴ Impulse = 20 N.sec
A ball of mass 0.5 kg falls from rest from a height of 0.9 m onto the surface of a viscous liquid. It then moves inside the liquid with a uniform velocity, covering 0.6 m in 0.5 sec. Find the magnitude of the impulse of the liquid on the ball at impact (g = 9.8 m/sec²).
Correct answer: (A) 1.5 N.sec
Explanation: Step 1: Velocity on reaching the surface: v₁² = 2 × 9.8 × 0.9 = 17.64, so v₁ = 4.2 m/sec downwards. Step 2: Velocity inside the liquid: v₂ = 0.6 ÷ 0.5 = 1.2 m/sec downwards. Step 3: Magnitude of impulse = m (v₁ − v₂) = 0.5 (4.2 − 1.2) = 1.5 N.sec (upwards).
A ball of mass 200 gm is projected vertically upwards with velocity 7 m/sec from a point 1.6 m below the ceiling of a room. It hits the ceiling and rebounds vertically downwards with velocity 2.8 m/sec. Find the magnitude of the impulse of the ceiling on the ball (g = 9.8 m/sec²).
Correct answer: (B) 1.4 N.sec
Explanation: Step 1: Velocity on reaching the ceiling: v² = 7² − 2 × 9.8 × 1.6 = 49 − 31.36 = 17.64, so v = 4.2 m/sec upwards. Step 2: Take downwards as positive: v₁ = −4.2 m/sec and v₂ = 2.8 m/sec. Step 3: Impulse = m (v₂ − v₁) = 0.2 (2.8 + 4.2) = 1.4 N.sec (downwards).
A ball of mass 0.5 kg falls from rest from a height of 2.5 m onto horizontal ground and rebounds vertically to a height of 0.9 m. The time of contact with the ground is 0.1 sec. Find the reaction of the ground on the ball during the impact. (g = 9.8 m/sec²)
Correct answer: (B) 60.9 N
Explanation: Speed before impact: v₁² = 2 × 9.8 × 2.5 ⇒ v₁ = 7 m/sec down; rebound speed: v₂² = 2 × 9.8 × 0.9 ⇒ v₂ = 4.2 m/sec up. Impulse = 0.5 × (4.2 + 7) = 5.6 N.sec ⇒ impulsive force F = 5.6 ÷ 0.1 = 56 N. For a body falling on the ground, reaction = F + w = 56 + 0.5 × 9.8 = 60.9 N. (56 N forgets the weight.)
A body of mass 2 kg moving with velocity (3 î + 4 ĵ) m/sec receives an impulse, after which it moves with the same speed in a direction perpendicular to its original direction. Find the magnitude of the impulse.
Correct answer: (B) 10√2 N.sec
Explanation: Step 1: The new velocity has speed 5 and is perpendicular to (3, 4), so it is (−4 î + 3 ĵ) or (4 î − 3 ĵ) m/sec. Step 2: Impulse = m(v − u) = 2(−7 î − ĵ) or 2(î − 7 ĵ); either way its magnitude is 2√50 = 10√2. ∴ 10√2 N.sec; 20 N.sec would be the impulse for reversing the direction of motion, not turning it through 90°.
The first 10 questions in every subject are free, each with an explanation.