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Transcript
State Examinations Commission – Physics Higher Level, 2004. Question 6 Define (i) force, (ii) momentum. (12) State Newton’s second law of motion. Hence, establish the relationship: force = mass × acceleration. (15) A pendulum bob of mass 10 g was raised to a height of 20 cm and allowed to swing so that it collided with a block of mass 8.0 g at rest on a bench, as shown. The bob stopped on impact and the block subsequently moved along the bench. Calculate (i) the velocity of the bob just before the collision; (ii) the velocity of the block immediately after the collision. (18) The block moved 2.0 m along the bench before stopping. What was the average horizontal force exerted on the block while travelling this distance? (11) (Acceleration due to gravity = 9.8 m s–2) _________________________________________________________________ Define (i) force, (ii) momentum. (12) A force is anything that causes a change in motion of a body. It is measured in Newtons. The momentum of a body is the product of its mass and velocity. It is measured is kg.m.s -1 . State Newton’s second law of motion. Hence, establish the relationship: force = mass × acceleration. (15) Newton’s second law states that if a resultant force F acts on some object, the rate of change of momentum of the object is proportional to that resultant force, and takes place in the direction of that force. mv − mu t m (v − u ) F∝ t F ∝ ma F∝ F = kma Where u is the initial and v the final velocity of the object after a time t during which the force F was acting. … Define 1N as that force which causes a 1kg mass to accelerate by 1ms -2 so, 1 = k. 1. 1 that is , k = 1 so , F = ma Chris Garvey 2004 A pendulum bob of mass 10 g was raised to a height of 20 cm and allowed to swing so that it collided with a block of mass 8.0 g at rest on a bench, as shown. The bob stopped on impact and the block subsequently moved along the bench. Calculate (i) the velocity of the bob just before the collision; Loss in potential energy of bob = gain in the kinetic energy of bob mgh = ½mv 2 gh = ½v2 2gh = v 2 √2gh = v √(2 x 9.8 x 0.2) = v v = 1.98ms -1 (ii) the velocity of the block immediately after the collision. (18) From principle of conservation of momentum, total momentum before collision = total after. 0.01 x 1.98 = 0.008 x v block v block = 2.48 ms -1 . The block moved 2.0 m along the bench before stopping. What was the average horizontal force exerted on the block while travelling this distance? (11) The deceleration of the block, a = (2.482 – 02 )/(2 x 2.0) a = 1.54 ms -2 … (v 2 = u 2 + 2as) Average horizontal force F = ma F = 0.008 x 1.54 F = 0.012N Chris Garvey 2004