Question
A toy rocket is made from a plastic bottle that contains some water.
Air is pumped into the vertical bottle until the pressure inside forces water and air out of the bottle. The bottle then travels vertically upwards.
The air–water mixture is called the propellant.
The variation with time of the vertical velocity of the bottle is shown.
The bottle reaches its highest point at time \(T_1\) on the graph and returns to the ground at time \(T_2\). The bottle then bounces. The motion of the bottle after the bounce is shown as a dashed line.
(a) Estimate, using the graph, the maximum height of the bottle. [3]
(b) Estimate the acceleration of the bottle when it is at its maximum height. [2]
(c) The bottle bounces when it returns to the ground.
(i) Calculate the fraction of the kinetic energy of the bottle that remains after the bounce. [2]
(ii) The mass of the bottle is 27g and it is in contact with the ground for 85ms.
Determine the average force exerted by the ground on the bottle. Give your answer to an appropriate number of significant figures. [3]
(d) After a second bounce, the bottle rotates about its centre of mass. The bottle rotates at 0.35 revolutions per second.
The centre of mass of the bottle is halfway between the base and the top of the bottle. Assume that the velocity of the centre of mass is zero. Calculate the linear speed of the top of the bottle. [3]
▶️Answer/Explanation
Ans:
ALTERNATIVE 1
Attempt to count squares \(\checkmark\) Area of one square found \(\checkmark\) 7.2 «m» (accept \(6.4-7.4 \mathrm{~m}\) )
ALTERNATIVE 2
b.Attempt to calculate gradient of line at \(t=1.2 \mathrm{~s} \checkmark\)
$
\left.\alpha-n 9.8 \ll \mathrm{m} \mathrm{s}^{-2} w \text { (accept } 9.6-10.0\right) \checkmark
$
Uses area equation for either triangle \(\checkmark\) Correct read offs for estimate of area of triangle \(\checkmark\) \(7.2 \mathrm{sm»} \mathrm{(accept} 6.4-7.4) \checkmark\)
c(i) Attempt to evaluate KE ratio as \(\left(\frac{v_{\text {final }}}{v_{\text {initial }}}\right)^2 \checkmark\)
$
«\left(\frac{4.5}{10}\right)^2=0.20 \text { OR } 20 \% \text { OR } \frac{1}{5}
$
c(ii) Attempt to use force \(=\) momentum change \(\div\) time \(\checkmark\)
$
\alpha=\frac{(4.5+10) \times 0.027}{85 \times 10^{-3}}=4.6 \text { ” }
$
$
\text { Force }=« 4.6+0.3 » 4.9 * \mathrm{~N} n \checkmark
$
Any answer to \(2 \mathrm{sf} \checkmark\)
d.ALTERNATIVE 1
$
\begin{aligned}
& \omega=2 \pi(0.35) \&=2.20 \mathrm{rad} \mathrm{s}^{-1} \rightsquigarrow \\
& \text { Use of } V=0.14 \omega \\
& 0.31 \ll \mathrm{m} \mathrm{s}^{-1} \rightsquigarrow
\end{aligned}
$
ALTERNATIVE 2
$
\begin{aligned}
& T=\frac{1}{0.35} \kappa=2.9 \mathrm{~s} » \\
& v=\frac{2 \pi(0.14)}{T} \text { oR } \frac{v^2}{0.14}=\frac{4 \pi^2(0.14)}{T^2} \\
& V=0.31 \mathrm{~km} \mathrm{~s}^{-1} \rightsquigarrow
\end{aligned}
$
e.Mass «leaving the bottle per second» will be larger for air-water \(\checkmark\) the momentum change/force is greater
Question
Part 2 Electric motor
An electric motor is used to raise a load.
a. Whilst being raised, the load accelerates uniformly upwards. The weight of the cable is negligible compared to the weight of the load.
(i) Draw a labelled free-body force diagram of the forces acting on the accelerating load. The dot below represents the load.
(ii) The load has a mass of 350 kg and it takes 6.5 s to raise it from rest through a height of 8.0 m.
Determine the tension in the cable as the load is being raised. [6]
(i) Calculate the power delivered to the load by the motor.
(ii) The current in the motor is 30 A. Estimate the efficiency of the motor.[4]
Answer/Explanation
Markscheme
a. (i) upward arrow labelled T/tension/force in cable and downward arrow labelled W/mg/weight/gravity force;{ (both needed)
tension arrow length >weight length;
(ii) \(a = \frac{{2s}}{{{t^2}}}\);
\(a = \left( {\frac{{2 \times 8.0}}{{{{6.5}^2}}} = } \right)0.38\left( {{\rm{m}}{{\rm{s}}^{ – 2}}} \right)\);
T=ma+mg or T=350(0.38+9.8);
3.6 kN;
Allow g=10 N kg-1 (same answer to 2 sf).
power \(\left( { = \frac{{24 \times {{10}^3}}}{{15}}} \right) = 1.6{\rm{kw}}\);
Allow g=10Nkg-1.
(ii) power input to motor=13.5 (kW);
efficiency=\(\left( {\frac{{1.6}}{{13.5}} = } \right)0.12\) or
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.
a. Draw the free-body diagram for the sledge at the position shown on the snow slope. [2]
Answer/Explanation
Markscheme
a. 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
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
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
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
«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
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»
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