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Lesson 3 questions – Newton’s Second Law ( /24)………………….%............... ALL 1 Fig 1.1 shows a car of mass 1200kg, pulling a caravan of mass 400kg along a horizontal road. Fig 1.2 Calculate, for the first 20s of the journey i) The acceleration of the car Gradient = 13/20 acceleration = …0.65…..ms-2 (2) ii) the resultant force acting on the car Force = ma 1200x0.65 resultant force on the car = 780…N (2) iii) the resultant force acting on the caravan. Force = ma 400x0.65 resultant force acting on the caravan = ……260…. N (2) MOST 2. A boy running at 8 m/s is stopped by a friend in 2s. If the mass of the boy is 40kg what is the force on him during stopping? 40x8 = Fx2 F = 160N ………160………………. N (2) 3. A ball of mass 20 g hits a vertical wall with a horizontal velocity of 12 m/s. If it rebounds with a velocity of 10 m/s in the opposite direction and is in contact with the wall for 0.05 s find: (a) the change in momentum of the ball; 0.02x22 = 0.44 Ns …………0.44 ……………. Kgm/s (2) (b) the force of the wall on the ball. 0.44/0.05 = 8.8 N …………8.8 ……………. N (2) 4. A parachutist lands on the ground. If his mass is 65 kg and he hits the ground at 4 m/s and stops in 0.5 s find: (a) his change of momentum; 65x4 = 260 Ns ………260 ………………. Kgm/s (2) (b) the force of the ground on him. 260/0.5 = 520 N ……… …520 ……………. N (2) 5. A second parachutist with the same mass lands with the same velocity but forgets to bend his knees and stops in 0.01 s. Find his change of momentum and the force of the ground on him. 260/0.01 = 26000 N! …………260 ……………. Kgm/s (2) ………26000 ………………. N (2) SOME 6. F=ma is a special case of Newton’s Second Law for a body that has a constant mass but changing velocity. a. Define a change in linear momentum, Δp in terms of Δt and F. ………………………………………………………………………………………………… …………………… Δp = F Δt …………………………………………………………………………… …………………………………………………………………………………………… (1) b. Define acceleration. ………………a= Δv/ Δt ………………………………………………………………………………… ………………………………………………………………………………………………… …………………………………………………………………………………………… (1) Show that F=ma is true for a body that has a constant mass but changing velocity using your definitions from (a) and (b). Since FΔt = Δp = Δ(mv); And acceleration is the change in velocity (Δv) in a change in time (Δt) or a = Δv/Δt; If an object of constant mass has a changing velocity we can write that: FΔt = m Δv If we rearrange this to make F the focus of the equation we can get: F = m Δv/Δt We have said that Δv/Δt = a and so substituting this in we now get F = ma which is a special case of Newton’s Law when mass remains constant and velocity changes. QED (2)