Question 1
1
An isolated spherical planet has a diameter of \(6.8 \times 10^{6} \mathrm{~m}\). Its mass of \(6.4 \times 10^{23} \mathrm{~kg}\) may be assumed to be a point mass at the centre of the planet.
structured3 marks
Question 1(b)
1(b)
A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m . Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone. Explain your working. J
Easystructured3 marks
Answer
\(\Delta E_{\mathrm{P}}=m g \Delta h\) because \(\Delta h \ll R\) (or \(1800 \mathrm{~m} \ll 3.4 \times 10^{6} \mathrm{~m}\) ) g is constant B1 \(\Delta E_{\mathrm{P}}=2.4 \times 3.7 \times 1800 \mathrm{C} 1\) (use of \(g=9.8 m s^{-2}\) max. 1 for explanation)
Question 1
1
An isolated spherical planet has a diameter of \(6.8 \times 10^{6} \mathrm{~m}\). Its mass of \(6.4 \times 10^{23} \mathrm{~kg}\) may be assumed to be a point mass at the centre of the planet.
structured3 marks
Question 1(b)
1(b)
A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m . Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone. Explain your working. J
Easystructured3 marks
Answer
\(\Delta E_{\mathrm{P}}=m g \Delta h\) because \(\Delta h \ll R\) (or \(1800 \mathrm{~m} \ll 3.4 \times 10^{6} \mathrm{~m}\) ) g is constant B1 (use of \(g=9.8 m s^{-2}\) max. 1 for explanation)
Question 1
1
A train of mass 600000 kg moves with a speed of \(100 \mathrm{~km} \mathrm{~h}^{-1}\). What is the order of magnitude of the kinetic energy of the train? \(10^{6} \mathrm{~J}\) \(10^{8} \mathrm{~J}\) \(10^{10} \mathrm{~J}\) \(10^{12} \mathrm{~J}\)
Mediummcq1 marks
Answer
B
Question 2
2
An Olympic athlete of mass 80 kg competes in a 100 m race. What is the best estimate of his mean kinetic energy during the race? \(4 \times 10^{2} \mathrm{~J}\) \(4 \times 10^{3} \mathrm{~J}\) \(4 \times 10^{4} \mathrm{~J}\) \(4 \times 10^{5} \mathrm{~J}\)
Mediummcq1 marks
Answer
B
Question 2
2
4 marks
Question 2(c)
2(c)
4 marks
Question 2(c)(i)
2(c)(i)
Calculate the maximum height above the ground of the ball in (b). maximum height = m
Mediumstructured2 marks
Answer
\[ \begin{array}{ll} \left(v^{2}=u^{2}+2 a s\right) 0=8.5^{2}+2(-9.81) s \text { or }\left(s=u t+\frac{1}{2} a t^{2}\right) s=8.5 \times 0.87+\frac{1}{2} \times(-9.81) \times 0.87^{2} \text { or }\left(s=v t-\frac{1}{2} a t^{2}\right) s=0-\frac{1}{2} \times(-9.81) \times 0.87^{2} \text { or }\left(s=\frac{1}{2}(u+v) t \text { or area under graph }\right) s=0.5 \times 8.5 \times 0.87 \end{array} \] C1 \(s=3.7 \mathrm{~m}\) A1
Question 2(c)(ii)
2(c)(ii)
The maximum gravitational potential energy of the ball above the ground is 22 J . Calculate the mass of the ball. mass = kg
Mediumstructured2 marks
Answer
\((\Delta) E=m g(\Delta) h\) C1 \[ \begin{aligned} m =22 /(9.81 \times 3.7) =0.61 \mathrm{~kg} \end{aligned} \] A1
Question 2
2
3 marks
Question 2(b)
2(b)
A force F acts on a mass m along a straight line for a distance s. The acceleration of the mass is a and the speed changes from an initial speed u to a final speed v.
structured3 marks
Question 2(b)(ii)
2(b)(ii)
Use your answer in (i) and an equation of motion to show that kinetic energy of a mass can be given by the expression
Mediumstructured3 marks
Answer
\(a s=\left(v^{2}-u^{2}\right) / 2\) any subject W=m a s hence \(W=m\left(v^{2}-u^{2}\right) / 2\) RHS represents terms of energy or with \(u=0 \mathrm{KE}=\frac{1}{2} m v^{2}\)
Question 4
4
An Olympic athlete of mass 80 kg competes in a 100 m race. What is the best estimate of his mean kinetic energy during the race? \(4 \times 10^{2} \mathrm{~J}\) \(4 \times 10^{3} \mathrm{~J}\) \(4 \times 10^{4} \mathrm{~J}\) \(4 \times 10^{5} \mathrm{~J}\)
Mediummcq1 marks
Answer
B
Question 2
2
2 marks
Question 2(c)
2(c)
The ball has a mass of 0.050 kg . It moves from A and reaches B after rebounding.
structured2 marks
Question 2(c)(i)
2(c)(i)
For this motion, calculate the change in 1. kinetic energy, change in kinetic energy = J [2] 2. gravitational potential energy. change in potential energy = J
Easystructured2 marks
Answer
1. kinetic energy at end is zero so \(\Delta \mathrm{KE}=1 / 2 m v^{2}\) or \(\Delta \mathrm{KE}=1 / 2 m u^{2}-1 / 2 m v^{2} \quad \mathrm{C} 1\) 2. final maximum height \(=(4.2)^{2} /(2 \times 9.8)=(0.9(\mathrm{~m}))\)
Question 2
2
A sphere is attached by a metal wire to the horizontal surface at the bottom of a river, as shown in Fig. 2.1. The sphere is fully submerged and in equilibrium, with the wire at an angle of \(68^{\circ}\) to the horizontal surface. The weight of the sphere is 32 N . The upthrust acting on the sphere is 280 N . The density of the water is \(1.0 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\). Assume that the force on the sphere due to the water flow is in a horizontal direction.
structured3 marks
Question 2(c)
2(c)
The centre of the sphere is initially at a height of 6.2 m above the horizontal surface. The speed of the water then increases, causing the sphere to move to a different position. This movement of the sphere causes its gravitational potential energy to decrease by 77 J . Calculate the final height of the centre of the sphere above the horizontal surface. height = m
Mediumstructured3 marks
Answer
\((\Delta) E=m g(\Delta) h\) or \((\Delta) E=W(\Delta) h\) C1 \((\Delta) h=(-) 77 / 32\) C1 \[ (\Delta) h=(-) 2.4 \] final height =6.2-2.4 \[ \text { = } 3.8 \mathrm{~m} \] A1
Question 2
2
A rigid uniform beam of weight W is connected to a fixed support by a hinge, as shown in Fig. 2.1. A compressed spring exerts a total force of 8.2 N vertically upwards on the horizontal beam. A block of weight 0.30 N rests on the beam. The right-hand end of the beam is connected to the ground by a string at an angle of \(30^{\circ}\) to the horizontal. The tension in the string is 4.8 N . The distances along the beam are shown in Fig. 2.1. The beam is in equilibrium. Assume that the hinge is frictionless.
structured14 marks
Question 2(c)
2(c)
The string is cut so that the spring extends upwards. This causes the beam to rotate and launch the block into the air. The block reaches its maximum height and then falls back to the ground. Fig. 2.2 shows part of the path of the block in the air shortly before it hits the horizontal ground. The block is at a height of 0.090 m above the ground when it passes through point A . The block has a kinetic energy of 0.044 J when it hits the ground at point B . Air resistance is negligible.
structured2 marks
Question 2(c)(i)
2(c)(i)
Calculate the decrease in the gravitational potential energy of the block for its movement from A to B . decrease in gravitational potential energy = J
Easystructured2 marks
Answer
\(((\Delta) E)=m g(\Delta) h\) or \(W(\Delta) h\) C1 \[ \begin{aligned} =0.30 \times 0.090 =0.027 \mathrm{~J} \end{aligned} \] A1