Forces and momentum for IB Physics HL

Reading cue. The area under the triangle is the impulse; 40 N is the peak value at one instant, not a force that acts throughout, so nothing on this graph may simply be multiplied by 0.20 s.

1. aState State the magnitude of the resultant force acting at t = 0.050 s.

   recall2 marks

   Check answer · 2 marks

   1. reads the rising straight line halfway between the origin and the peak
   2. F = 20 N

2. bSketch The object has mass 0.50 kg. Sketch the acceleration-time graph for the same 0.20 s, marking values on both axes.

   routine3 marks

   Check answer · 3 marks

   1. same triangular shape: straight rise from the origin, straight fall to zero at t = 0.20 s
   2. peak acceleration = 40 / 0.50 = 80 m s⁻² at t = 0.10 s
   3. axes labelled a / m s⁻² and t / s with 80, 0.10 and 0.20 marked

3. cDetermine Determine the impulse delivered during the first 0.10 s, and hence determine the time at which the object, starting from rest, is moving at half of its final speed.

   demanding3 marks

   Check answer · 3 marks

   1. impulse to t = 0.10 s = ½ × 0.10 × 40 = 2.0 N s
   2. total impulse = ½ × 0.20 × 40 = 4.0 N s, so exactly half the momentum has been delivered by the peak
   3. mass is constant and the object started from rest, so half the momentum is half the speed: t = 0.10 s

4. dDetermine The same force pulse is now applied to a 0.50 kg object already moving at 8.0 m s⁻¹ in the direction opposite to the force. Determine its velocity at t = 0.20 s, and explain why the same area under the graph still applies.

   top of the paper4 marks

   Check answer · 4 marks

   1. area under the graph = ½ × 0.20 × 40 = 4.0 N s
   2. impulse gives the change in momentum, so Δv = 4.0 / 0.50 = 8.0 m s⁻¹ in the direction of the force
   3. taking the force direction as positive, v = −8.0 + 8.0 = 0, so the object is momentarily at rest
   4. the area fixes the change in velocity only, and is independent of the velocity the object started with

Reading cue. The 12 m s⁻¹ arrow is horizontal and is the whole initial velocity, so the ball leaves the roof with zero vertical speed and 25 m is a free-fall drop, not a length of the dashed curve.

1. aState State the horizontal component of the ball's velocity at the instant it reaches the ground, and give a reason.

   recall2 marks

   Check answer · 2 marks

   1. 12 m s⁻¹
   2. no horizontal force acts while air resistance is negligible, so the horizontal component is unchanged throughout

2. bDetermine Determine the horizontal distance from the foot of the building to the point where the dashed path meets the ground.

   routine3 marks

   Check answer · 3 marks

   1. vertical: 25 = ½ × 9.81 × t², so t = 2.26 s
   2. horizontal motion is at constant velocity: x = 12 × 2.26
   3. x = 27 m

3. cDetermine Determine the angle below the horizontal at which the dashed path meets the ground.

   demanding3 marks

   Check answer · 3 marks

   1. vertical component on landing: v_y = 9.81 × 2.26 = 22.1 m s⁻¹
   2. horizontal component is still 12 m s⁻¹, so tan θ = 22.1 / 12 = 1.85
   3. θ = 62° below the horizontal

4. dDiscuss A second ball is thrown horizontally from the same point at 24 m s⁻¹. Discuss how its time of flight, its landing distance and its landing angle each compare with those of the first ball.

   top of the paper4 marks

   Check answer · 4 marks

   1. the vertical motion is unaffected by the horizontal velocity, so the time of flight is the same, 2.26 s
   2. the landing distance doubles to 54 m, since x = ut with the same t
   3. the vertical component on landing is still 22.1 m s⁻¹, so tan θ = 22.1 / 24 and θ = 43°
   4. concludes that throwing harder flattens the path and moves the landing point out, but does not keep the ball in the air any longer