| Date | November 2017 | Marks available | 3 | Reference code | 17N.2.SL.TZ0.1 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 0 |
| Command term | Outline | Question number | 1 | Adapted from | N/A |
Question
A girl on a sledge is moving down a snow slope at a uniform speed.
The sledge, without the girl on it, now travels up a snow slope that makes an angle of 6.5˚ to the horizontal. At the start of the slope, the speed of the sledge is 4.2 m s–1. The coefficient of dynamic friction of the sledge on the snow is 0.11.
Draw the free-body diagram for the sledge at the position shown on the snow slope.
After leaving the snow slope, the girl on the sledge moves over a horizontal region of snow. Explain, with reference to the physical origin of the forces, why the vertical forces on the girl must be in equilibrium as she moves over the horizontal region.
When the sledge is moving on the horizontal region of the snow, the girl jumps off the sledge. The girl has no horizontal velocity after the jump. The velocity of the sledge immediately after the girl jumps off is 4.2 m s–1. The mass of the girl is 55 kg and the mass of the sledge is 5.5 kg. Calculate the speed of the sledge immediately before the girl jumps from it.
The girl chooses to jump so that she lands on loosely-packed snow rather than frozen ice. Outline why she chooses to land on the snow.
Show that the acceleration of the sledge is about –2 m s–2.
Calculate the distance along the slope at which the sledge stops moving. Assume that the coefficient of dynamic friction is constant.
The coefficient of static friction between the sledge and the snow is 0.14. Outline, with a calculation, the subsequent motion of the sledge.
Markscheme
arrow vertically downwards labelled weight «of sledge and/or girl»/W/mg/gravitational force/Fg/Fgravitational AND arrow perpendicular to the snow slope labelled reaction force/R/normal contact force/N/FN
friction force/F/f acting up slope «perpendicular to reaction force»
Do not allow G/g/“gravity”.
Do not award MP1 if a “driving force” is included.
Allow components of weight if correctly labelled.
Ignore point of application or shape of object.
Ignore “air resistance”.
Ignore any reference to “push of feet on sledge”.
Do not award MP2 for forces on sledge on horizontal ground
The arrows should contact the object
gravitational force/weight from the Earth «downwards»
reaction force from the sledge/snow/ground «upwards»
no vertical acceleration/remains in contact with the ground/does not move vertically as there is no resultant vertical force
Allow naming of forces as in (a)
Allow vertical forces are balanced/equal in magnitude/cancel out
mention of conservation of momentum
OR
5.5 x 4.2 = (55 + 5.5) «v»
0.38 «m s–1»
Allow p=p′ or other algebraically equivalent statement
Award [0] for answers based on energy
same change in momentum/impulse
the time taken «to stop» would be greater «with the snow»
\(F = \frac{{\Delta p}}{{\Delta t}}\) therefore F is smaller «with the snow»
OR
force is proportional to rate of change of momentum therefore F is smaller «with the snow»
Allow reverse argument for ice
«friction force down slope» = μmg cos(6.5) = «5.9 N»
«component of weight down slope» = mg sin(6.5) «= 6.1 N»
«so a = \(\frac{F}{m}\)» acceleration = \(\frac{{12}}{{5.5}}\) = 2.2 «m s–2»
Ignore negative signs
Allow use of g = 10 m s–2
correct use of kinematics equation
distance = 4.4 or 4.0 «m»
Alternative 2
KE lost=work done against friction + GPE
distance = 4.4 or 4.0 «m»
Allow ECF from (e)(i)
Allow [1 max] for GPE missing leading to 8.2 «m»
calculates a maximum value for the frictional force = «μR=» 7.5 «N»
sledge will not move as the maximum static friction force is greater than the component of weight down the slope
Allow correct conclusion from incorrect MP1
Allow 7.5 > 6.1 so will not move
