M2 June 2013 Q6

EdexcelOld spec11 marksProjectiles

Figure 4: ball projected from A, 8 m above horizontal ground, with speed u m s^-1 at theta degrees above the horizontal, landing at B — Figure 4

A ball is projected from a point \(A\) which is 8 m above horizontal ground as shown in Figure 4. The ball is projected with speed \(u\) m s\(^{-1}\) at an angle \(\theta^\circ\) above the horizontal. The ball moves freely under gravity and hits the ground at the point \(B\). The speed of the ball immediately before it hits the ground is \(2u\) m s\(^{-1}\).

(a) By considering energy, find the value of \(u\). (5)

The time taken for the ball to move from \(A\) to \(B\) is 2 seconds. Find

(b) the value of \(\theta\), (4)

Mark scheme (a)

Scheme	Marks
Conservation of energy:	M1
\(\dfrac{1}{2}mu^2 + mg \times 8 = \dfrac{1}{2}m(2u)^2\)	A2 -1ee
\(mu^2 + 16mg = 4mu^2\)
\(16mg = 3mu^2,\ \ \ u = \sqrt{\dfrac{16g}{3}}\)	DM1
\(u = 7.2\)	A1
	(5)

Notes

M1 Energy equation must contain the correct terms, but condone sign error.

A2 -1ee Correct unsimplified

DM1 Solve for \(u\)

A1 Accept 7.23. Accept \(\sqrt{\dfrac{16g}{3}}\)

Mark scheme (b)

Scheme	Marks
Vertical distance: \(-8 = u\sin\theta \times 2 - \dfrac{g}{2} \times 4\)	M1 A2 -1ee
\(\sin\theta = \dfrac{2g - 8}{2u} = 0.802\ldots\)
\(\theta = 53.3^\circ\)	A1
	(4)

Notes

M1 Condone sign errors or trig error. \(u\) must be resolved.

A2 -1ee Correct equation for their \(u\).

A1 or \(53^\circ\)

Scheme	Marks
Min speed at max height, i.e. \(u\cos\theta\)	M1
\(= 4.3\) (m s\(^{-1}\))	A1
	(2)
	(11 marks)