Physics 1st Ch 03 Dynamics

Physics - First Paper Chapter 3: Dynamics Selected Questions and Answers (HSC English Version) Chapter 3: Dynamics B. Creative Questions 1. A train started moving from Kamalapur Railway Station at uniform acceleration with a speed of 0.5 m/s. At the same time a hare started running at uniform speed of 5 m/s parallel to the train. At one stage the train crosses the hare. (a) What is instantaneous speed? (b) Why a body moving at uniform speed does not have acceleration? (c) After how many metres travelled by the hare the train will cross the hare? Solution : Given: Train initial speed u = 0.5 m/s, acceleration a (unknown, but this is the train speed); hare speed vh = 5 m/s. Wait - re-reading: the train starts with uniform acceleration of a = 0.5 m/s2 (the value is the acceleration, starting from rest). Hare: uniform speed 5 m/s. Train crosses hare when strain = share (starting from same point simultaneously). strain = 1 2at2 = 1 2(0.5)t2 = 0.25t2 share = 5t Setting equal: 0.25t2 = 5t =⇒t(0.25t −5) = 0 =⇒t = 20 s (non-trivial). Distance by hare: share = 5 × 20 = 100 m. The train crosses the hare after the hare has travelled 100 m. (d) Does the event mentioned in the stimulus support t

(action), and the track exerts an equal forward reaction on the train (reaction), providing the net forward force that accelerates the train. Without this action-reaction pair, the train could not accelerate at a = 0.5 m/s2. Mathematically, at the moment of overtaking (t = 20 s): vtrain = at = 0.5 × 20 = 10 m/s > vhare = 5 m/s. The train is moving faster, confirming that its net forward force (from the reaction of the track) has been effective throughout. Newton’s third law is thus implicitly supported by the mechanics of the event. 2. Tania dropped a body from the peak of a tower of 180 m height. At the same time her friend Joyita threw a body vertically upward at a speed of 60 m/s. The thrown body returned to the ground after reaching the maximum height. (a) What is uniform speed? (b) Why the acceleration of a falling body is uniform acceleration? (c) Where and when the two bodies in the stimulus will meet? Solution : Take upward as positive; origin at ground. Let t = time after release. Dropped body (from tower top, h0 = 180 m, u = 0, a = −g = −9.8): y1 = 180 −1 2(9.8)t2 Thrown body (from ground, u = 60, a = −g = −9.8): y2 = 60t −1 2(9.8)t2 Meeting: y1 = y2: 180 −1 2(9.8)t2 = 60

Substituting back: H = utH −1 2gt2 H = u · u g −1 2g · u2 g2 = u2 g −u2 2g = u2 2g . Confirmed. Physical interpretation: at the highest point all kinetic energy has been converted to potential energy: 1 2mu2 = mgh ⇒h = u2/(2g). 3. At one stage of a football match between France and Germany football player Zidan of France kicked a ball from a distance 10.97 m of the goal post and the ball rushed directly towards the mid point of the goal post at an angle of 30◦with velocity of 14 m/s. The goal keeper standing 1 m in front of the goal post of height 2.44 m could catch the ball from a maximum height of 2.2 m. (a) What is mean velocity? (b) How the motion of a boat can be increased in case of towing a boat - explain. (c) Calculate the maximum height of the ball. Solution : Given: v0 = 14 m/s, θ = 30◦, g = 9.8 m/s2. Vertical component: vy = v0 sin θ = 14 sin 30◦= 7 m/s. Maximum height: H = v2 y 2g = 72 2 × 9.8 = 49 19.6 H ≈2.5 m (d) Whether there will be goal mentioned in the stimulus or not - explain mathematically. Answer: (a) Mean (average) velocity is the total displacement divided by the total time taken: ¯v = ∆⃗r/∆t. It is a vector in the direction of displacement. It may differ f