M1 June 2018 Q4

EdexcelOld spec13 marksSUVAT Time-Series Graphs

4. A ball of mass 0.2 kg is projected vertically downwards with speed \(U\) m s\(^{-1}\) from a point \(A\) which is 2.5 m above horizontal ground. The ball hits the ground. Immediately after hitting the ground, the ball rebounds vertically with a speed of 10 m s\(^{-1}\). The ball receives an impulse of magnitude 7 N s in its impact with the ground. By modelling the ball as a particle and ignoring air resistance, find

(a) the value of \(U\). (6)

After hitting the ground, the ball moves vertically upwards and passes through a point \(B\) which is 1 m above the ground.

(b) Find the time between the instant when the ball hits the ground and the instant when the ball first passes through \(B\). (4)

(c) Sketch a velocity-time graph for the motion of the ball from when it was projected from \(A\) to when it first passes through \(B\). (You need not make any further calculations to draw this sketch.) (3)

Mark scheme (a)

Scheme	Marks
\(V^2 = U^2 + 2g \times 2.5\)	M1A1
Eliminate \(V\) and solve for \(U\)	A1 (DM1)
\(7 = 0.2(10 - -V)\)	M1A1
\(U = 24\)	A1
	(6)

Notes

First M1 for complete method, using suvat, to find equation in \(U\) and \(V\) only

First A1 for a correct equation

Second A1 – treat as third DM1, dependent on the other two M’s, for eliminating \(V\) and solving for \(U\)

Second M1 for using Impulse = Change in Momentum of ball (must have 0.2 in both terms and be using 10 as one of the velocities) (M0 if clearly adding momenta or if \(g\) is included) but condone sign errors.

Third A1 for a correct equation, 7 and 10 must have the same sign but equation may have \(V\) instead of \(-V\)

Fourth A1 for \(U = 24\) (must appear here)

N.B. If they use \(U\) instead of \(V\) in the impulse-momentum equation, can score max M1A0/6 for part (a).

N.B. If they go from \(V^2 = U^2 + 49\) to \(V = U + 7\), can score max 5/6

Mark scheme (b)

Scheme	Marks
\(1 = 10t - 4.9t^2\) OR e.g. \(v^2 = 10^2 - 2 \times 9.8 \times 1\) and \(v = 10 - 9.8t\)
\(1 = 10t - 4.9t^2\) to give \(\sqrt{80.4} = 10 - 9.8t\)	M1 A1
\(t = \dfrac{10 \pm \sqrt{100 - 19.6}}{9.8}\) so \(t = \dfrac{10 - \sqrt{10^2 - 2 \times 9.8 \times 1}}{9.8}\)	DM1
\(t = 0.11\) s or 0.105 s	A1
	(4)

Notes

First M1 for complete method, using one or more suvat formulae, to produce an equation in \(t\) only using \(s = 1\) or \(-1\)

First A1 for a correct equation in \(t\) only

Second DM1, dependent on first M1, for solving their equation (this mark can be implied by a correct answer)

Second A1 for either 0.105 (s) or 0.11 (s) (must be only ONE answer)

ALTERNATIVE

ALTERNATIVE : “the instant when the ball first passes through \(B\)” is taken to be when the ball is on the way down from \(A\).
\(s = vt - \dfrac{1}{2}at^2\) OR \(v_B^2 = 24^2 + 2 \times 9.8 \times 1.5\) and \(25 = v_B + 9.8t\)
\(1 = 25t - 4.9t^2\) to give \(25 = \sqrt{605.4} + 9.8t\)	M1 A1
\(t = \dfrac{25 \pm \sqrt{625 - 19.6}}{9.8}\) so \(t = \dfrac{25 - \sqrt{625 - 19.6}}{9.8}\)	DM1
\(t = 0.040\) (s) or 0.0403 (s) or 0.04 (s) (must only be ONE answer)	A1

First M1 for complete method, using one or more suvat formulae, to produce an equation in \(t\) only using \(s = 1\) or \(-1\)

First A1 for a correct equation in \(t\) only

Second DM1, dependent on first M1, for solving their equation (this mark can be implied by a correct answer)

Second A1 \(t = 0.040\) (s) or 0.0403 (s)

Velocity-time graph: line of positive gradient from 24 on the v-axis, then a parallel line starting at -10 at the same time — Mark scheme (c)

Velocity-time graph: single line of positive gradient starting at 24 on the v-axis — Mark scheme (c)

Scheme	Marks
	B1ft 1^st line
	B1 2^nd line
	B1 ,-10
	(3)
	(13 marks)