Specimen M1 Paper Mark Scheme

1. coefficient of friction = tan (22) [1]
  0.404 (3sf) [1]
2. angle = arctan(2*coefficient of friction) [1]
  38.9 degrees (3sf) [1, follow through]
3. The coefficient of friction does not depend on the value of the constant acceleration the system is under, or the number of points at which the object rests on the surface. [1]
  0.808 (3sf) [1, follow through]
1. Work = Force Distance = 250J. [1]
  Work = Energy imparted to arrow = Kinetic energy of arrow = 0.5 m vv = 250J. [1]
  v = 50m/s.
  gt = v where t is the time for maximum height. [1]
  t = 5.10 (3sf) s.
  height = 0.5 g tt. [1]
  height = 255 (3sf) m. [2, 1 follow through]
1. Angle of rope to horizontal = 45 degrees.
  Force required at point where rope is attached to drawbridge = (2.5m / 5m) 500kg g = 250kg g [2]
  Tension sin(45) = 250kg g [1]
  Tension = 3460 (3sf) N. [2]
2. Tension sin(45-d/2) = 500kg g cos(d). [1]
  Tension = 500kg g cos(d)/sin(45-d/2)
  More. [2]
At the bottom, the reeve's horizontal velocity (u) can be calculated. The time, t, of his jump is found from g(t/2) = 1 m/s [1].
t = 0.204s (3sf).
t u = 3m.
u = 14.7m/s. [1]
u / cos( angle ) = v, where v is the velocity of the reeve at the bottom of the hill.
The acceleration of the reeve down the hill is g sin(angle), so v = g sin(angle) t. [1]
u = g t sin(angle) cos(angle)
u = g t .5 sin( 2 angle )
angle = 1.43 degrees (3sf). [2]
Energy is conserved, so 0.5 m vv = m g h.
h = 11.0m (3sf). [2]
F t = change in momentum. [1]
1000N 0.2s = 200Ns
200Ns / 100kg = 2 m/s = v. [1]
v = at = gt.
t = 0.204s (3sf).
0.5g tt = length of drop.
length of drop = 0.204m (3sf). [total 2, 1 follow through].
Friar Tuck as particle (implying, no air resistance, etc). Constant gravitational acceleration.
1. First, the centre of mass of the arrowhead from the base of the shaft. The height of the triangle is sqrt( 4cm*4cm - 1cm*1cm ) = 3.87cm (3sf) [1].
  The centre of mass of the triangle is one third of the way along this line. [1]
  Centre of mass of arrowhead is 1.29cm (3sf) from base of shaft. [1 follow through]
  Centre of mass of shaft is half way down it. [1]
  Centre of mass of system is along the line of the shaft.
  d = ((40cm / 2) + 1.29cm) m)/(2 m) where d is the distance of the centre of mass from the centre of the shaft.
  10.6cm (3sf). [total 2, 1 follow through]
2. 1. The system has to at rest when pivoted at its centre of mass (assuming constant gravitational field), so 0Nm. [1]
  2. If a diagram is drawn, *deduct* 2 marks for this question. The system has to at rest when pivoted at its centre of mass (assuming constant gravitational field), so 0Nm. [2]
1. 1. (50m/s sin(60) p - .5 g pp) i + 50m/s cos(60) p j. [2]
  2. (75m/s sin(45) q - .5 g qq) i + 75m/s cos(45) q j. [2]
2. The i components are equal and the j components are equal. [1]
  Let p = q + t.
  t = (75m/s cos(45) - 50m/s cos(60))q / (50m/s cos(60)).
  t = 1.12q (3sf).
  50m/s sin(60) 2.12q - .5 g 4.5qq = 75m/s sin(45) q - .5 g qq.
  38.8 q = 26.95 qq.
  q = 0 or 1.44s (3sf)
  t is 1.61s.
3. Subtract a small amount from 75 m/s and do the question again.
  More.
1. Let the force of the horse on the cart be F.
  m a = F - m g sin(30). [2]
  F = m a + m g sin(30).
  m = (300kg - 175.5 kg).
  Fv = P. [1]
  P = 735W (3sf).
  European (see data given at top of paper). [2]
2. P = m a v + m v g sin(angle).
  arcsin( (P - m a v) / (m g v) ) = angle.
  arcsin( (P - 124.5W) / (1220.1W) ) = angle. [2]
1. The tension is proportional to the weight, which is proportional to the mass. Half the tension therefore means half the mass, so the 10kg can be raised at the same acceleration as the 20kg before, that is, 5m/s/s. [2]
2. m (a + 9.8m/s/s) = 148N. [1]
  m = 148N / (a + 9.8m/s/s).
  0.5 a tt = 10m. [1]
  t = sqrt( 20m / a )
  m / t = 148N sqrt(a) / ((a + 9.8m/s/s) sqrt( 20m ))
  Let r = m / t.
  dr/da = k ( (a + 9.8m/s/s) .5 1/sqrt(a) - sqrt(a) ) / ( aa + 19.6a + 96.04 )
  dr/da = 0 when a = 9.8m/ss. This is a maximum.
  a = 9.8m/ss [4], m = 7.55kg [1] (3sf). Assumptions: constant acceleration, no air resistance, no vibration of bucket at end of rope.
The peasant is deindividualised. It could be argued that one should obtain his or her consent before performing ethically dubious thought experiments on him or her.