M2 June 2018 Q1

EdexcelOld spec8 marksPower

1. A truck of mass 750 kg is moving with constant speed \(v\) m s\(^{-1}\) down a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{3}{49}\). The resistance to motion of the truck is modelled as a constant force of magnitude 1200 N. The engine of the truck is working at a constant rate of 9 kW.

(a) Find the value of \(v\). (4)

On another occasion the truck is moving up the same straight road. The resistance to motion of the truck from non-gravitational forces is modelled as a constant force of magnitude 1200 N. The engine of the truck is working at a constant rate of 9 kW.

(b) Find the acceleration of the truck at the instant when it is moving with speed 4.5 m s\(^{-1}\). (4)

Force diagram: truck on slope at angle theta moving down at v m/s with forces R, 750g, F down the slope and 1200 N up the slope — Mark scheme (a)

Force diagram: truck on slope moving up at 4.5 m/s with forces R, 750g, F up the slope and 1200 N down the slope — Mark scheme (b)

Scheme	Marks
Motion down the plane:	M1
\(F + 750g\sin\theta = 1200\)	A1
Use of \(P = Fv\) : \(\ F = \dfrac{9000}{v}\)	B1
\(\dfrac{9000}{v} + 750g \times \dfrac{3}{49} = 1200\)
\(v = \dfrac{9000}{750} = 12\)	A1
	(4)

Scheme	Marks
\(F = ma\) : \(\ F - (750g\sin\theta + 1200) = 750a\)	M1
\(\dfrac{9000}{4.5} - \left(750g \times \dfrac{3}{49} + 1200\right) = 750a\)	A1 A1
\(a = 0.47\ (0.467)\) (m s\(^{-2}\))	A1
	(4)
	(8 marks)

M2 June 2018 Q1

M2 June 2018 Q1

Notes

Notes