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Question 1: Linear Acceleration Please remember to photocopy 4 pages onto one sheet by going A3→A4 and using back to back on the photocopier
Page
Introduction 2
Introductory questions taken from Physics Papers 3
Solutions to above 4
Ordinary Level Exam Questions – Worked Solutions 5
Ordinary Level Exam Questions 8
Answers to Ordinary Level Exam Questions 10
Higher Level Introduction to vertical motion 12 Introductory questions taken from Physics Papers 13
Solutions to above 14
Higher Level Applied Maths Exam Questions Vertical Motion
Common Initial Velocity 17
F = ma (^) 19
Multi-stage Problems 20
General Questions 26
Guide to answering individual higher level exam questions 2009 – 1995 28
Other miscellaneous points 33
************* Marking Schemes / Solutions to be provided separately ***************
Acceleration is the rate of change of velocity with respect to time.* The unit of acceleration is the metre per second squared (m s-2, or m/s^2 ).
Equations of Motion* When an object (with initial velocity u ) moves in a straight line with constant acceleration a , its displacement s from its starting point, and its final velocity v , change with time t. Note that both v and u are instantaneous velocities.
The following equations tell us how these quantities are related:
v = final velocity u = initial velocity a = acceleration s= displacement (not distance) t = time
Procedure for solving problems using equations of motion.
- Write down v, u, a, s and t underneath each other on the left hand side of the page, filling in the quantities you know, and put a question mark beside the quantity you are looking for.
- Write down the three equations of motion every time.
- Decide which of the three equations has only one unknown in it.
- Substitute in the known values in to this equation and solve to find the unknown.
Velocity – Time graphs (for an object travelling with constant acceleration) If a graph is drawn of Velocity (y-axis) against Time (x-axis), the slope of the graph is the acceleration of the object.
Note that the area under each section of the graph corresponds to the distance travelled in that section
v = u + at
s = ut + ½ at
2
v
2
= u
2
+ 2as
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
Solutions
v = u + at ⇒ a = ( v – u ) ÷ t ⇒ a = (30 – 10) ÷ 5 ⇒ a = 4 m s-2.
v = u + at ⇒ 0 = 60 + a (120) ⇒ a = - 0.5 m s-
(i) See diagram (ii) v = u + at ⇒ 28 = 0 + a (4) ⇒ a = 7 m s-
(i)
(ii) 12 m s-1. (iii) v = u + at but u = 0 ⇒ a = v/t = 20/10 = 2 m s-2. (iv) v = s/t ⇒ s = vt = 20 × 5 = 100 m
- v 2 = u 2 + 2as 0 = 25 +2(-1.47) s or s = 213 m
- v^2 = u^2 + 2as ⇒ (12.2)^2 = 0 +2a(25) a = 2.98 m s–
(i) v = u + at 50 = 0 + 0.5t t = 50/0.5 = 100 s (ii) s = ut + ½ at^2 (but a = 0) s = 50 × (90×60) = 270000 m (iii) v^2 = u^2 + 2as 0 = 50^2 + 2a(500) a = −2500/1000 = − 2.5 m s-
Exam Questions Ordinary Level – Worked Solutions
2009 OL 3 points p , q and r lie on a straight level road. Two cars, A and B, are moving towards each other on the road. Car A passes p with speed 3 m/s and uniform acceleration of 2 m/s^2 and at the same instant car B passes r with speed 5 m/s and uniform acceleration of 4 m/s^2. A and B pass each other at q seven seconds later. Find (i) the speed of car A and the speed of car B at q. (ii) | pq | and | rq |, the distances A and B have moved in these 7 s. (iii)Car A stops accelerating at q and continues on to r at uniform speed. Find, correct to one place of decimals, the total time for car A to travel from p to r.
Solution (i) v = u + at vA = 3 + 2(7) vA =17 m/s
v = u + at vB = 5 + 4(7) vB =33 m/s
(ii) s = ut + ½ at^2 sA = 3(7) + ½ 2(49) sA =70 m
s = ut + ½ at^2 sB = 5(7) + ½ 4(49) sB =133 m
(iii)s = ut + ½ at^2 133 = 17(t) + 0 t = 7.8 s
total time = 14.8 s
2007 OL
A car travels from p to q along a straight level road. It starts from rest at p and accelerates uniformly for 5 seconds to a speed of 15 m/s. It then moves at a constant speed of 15 m/s for 20 seconds. Finally the car decelerates uniformly from 15 m/s to rest at q in 3 seconds. (i) Draw a speed-time graph of the motion of the car from p to q. (ii) Find the uniform acceleration of the car. (iii)Find the uniform deceleration of the car. (iv) Find | pq |, the distance from p to q. (v) Find the speed of the car when it is 13.5 metres from p.
Solution (i)
(ii) v = u + at 15 = 0 + 5 a a = 3 m s-
(iii)v = u + at 0 = 15 + 3a a = - 5 deceleration is 5 m s-
(iv) distance = ½ (5)(15) + (20)(15) + ½ (3)(15) = 37.5 + 300 + 22. = 360 m
(v) v^2 = u^2 + 2as = 0 + 2(3)(13.5) = 81 m v = 9 m s-
2010 OL
A car travels along a straight level road. It passes a point P at a speed of 12 m s-1^ and accelerates uniformly for 6 seconds to a speed of 30 m s-1. It then travels at a constant speed of 30 m s-1^ for 15 seconds. Finally the car decelerates uniformly from 30 m s-1^ to rest at a point Q. The car travels 45 metres while decelerating. Find (i) the acceleration (ii) the deceleration (iii)|PQ|, the distance from P to Q (iv) the average speed of the car as it travels from P to Q.
2006 OL
A car travels along a straight level road. It passes a point p at a speed of 10 m/s and accelerates uniformly for 5 seconds to a speed of 30 m/s. It then moves at a constant speed of 30 m/s for 9 seconds. Finally the car decelerates uniformly from 30 m/s to rest at point q in 6 seconds. Find (i) the acceleration (ii) the deceleration (iii) pq , the distance from p to q (iv) the average speed of the car as it travels from p to q.
2005 OL
A particle travels from p to q in a straight line. It starts from rest at p and accelerates uniformly to its maximum speed of 20 m/s in 10 seconds. The particle maintains this speed of 20 m/s for 15 seconds before decelerating uniformly to rest at q in a further 20 seconds. (i) Draw a speed-time graph of the motion of the particle from p to q. (ii) Find the uniform acceleration of the particle. (iii)Find the uniform deceleration of the particle. (iv) Find pq , the distance from p to q. (v) Find the average speed of the particle as it moves from p to q, giving your answer in the form a/b where a , b ∈ N.
2004 OL
Three points a, b and c, lie on a straight level road such that ab=bc= 100 m. A car, travelling with uniform retardation, passes point a with a speed of 20 m/s and passes point b with a speed of 15 m/s. (i) Find the uniform retardation of the car. (ii) Find the time it takes the car to travel from a to b, giving your answer as a fraction. (iii)Find the speed of the car as it passes c, giving your answer in the form p q , where p, q ∈ N. (iv) How much further, after passing c, will the car travel before coming to rest? Give your answer to the nearest metre.
Answers to Ordinary Level Exam Questions 2010 (i) a = 3 m s- (ii) a = -10 m s- (iii) PQ = 621 m (iv) Average speed = 25.875 m s-
2009 (i) VA = 17 m s-1, VB = 33 m s- (ii) SA = 70 m, SB = 133 m (iii)t = 14.8 s
2008 (i) Retardation = 2.5 m s- (ii) t = 4 s (iii)s = 80 m (iv) v = 14.1 m s-
2007 (i) (ii) a = 3 m s- (iii)a = - 5 m s- (iv) s = 360 m (v) v = 9 m s-
2006 (i) Acceleration = 4 m s- (ii) Deceleration = 5 m s- (iii)Distance = 460 m (iv) Average speed = 23 m s-
2005 (i) (ii) a = 2 m s- (iii)Deceleration = 1 m s- (iv) s = 600 m (v) average speed = 40/3 m s-
2004 (i) Retardation = - 0.875 m s- (ii) t = 40/7 s (iii)v = 5√2 m s- (iv) s = 29 m
2003 (i) (ii) a = 1 m s- (iii)a = -4 m s- (iv) s = 80 m (v) v = 40/3 m s-
(i) a = 2 m s- (ii) t = 40 s (iii)t = 75s
(i) VA = 22 m s- VB = 31 m s- (ii) SA = 120 m SB = 160 m (iii)u = 14 m s-
2000 (i) (ii) Acceleration = 2.5 m s- (iii)Deceleration = 5 m s- (iv) Distance = 715 m (v) time = 36.1 s
Questions taken from Leaving Cert Physics Exam Papers
- [2005] A basketball which was resting on a hoop falls to the ground 3.05 m below. What is the maximum velocity of the ball as it falls?
- [2006 OL] An astronaut drops an object from a height of 1.6 m above the surface of the moon and the object takes 1.4 s to fall. Calculate the acceleration due to gravity on the surface of the moon.
- [2003 OL] (i) An astronaut is on the surface of the moon, where the acceleration due to gravity is 1.6 m s–2. The astronaut throws a stone straight up from the surface of the moon with an initial speed of 25 m s–1. Describe how the speed of the stone changes as it reaches its highest point. (ii) Calculate the highest point reached by the stone. (iii) Calculate how high the astronaut can throw the same stone with the same initial speed of 25 m s– when on the surface of the earth, where the acceleration due to gravity is 9.8 m s–2.
- [2003] A skydiver falls from an aircraft that is flying horizontally. He reaches a constant speed of 50 m s–1^ after falling through a height of 1500 m. Calculate the average vertical acceleration of the skydiver.
- [2006] The student releases the ball when is it at A, which is 130 cm above the ground, and the ball travels vertically upwards at a velocity of 7 m s-1. Calculate the maximum height, above the ground, the ball will reach.
- [2006] The student releases the ball when is it at A, which is 130 cm above the ground, and the ball travels vertically upwards at a velocity of 7 m s-1. Calculate the time taken for the ball to hit the ground after its release from A.
Solutions
v^2 = u^2 +2as ⇒ v^2 = 0 + 2(9.8)(3.05) ⇒ v^2 = 59. v = 7.73 m s-
s = 1.6 m, t = 1.4 s, u = 0. Substitute into the equation s = ut + ½ at^2 to get a = 1.6 m s-2.
(i) It slows? (ii) v^2 = u^2 + 2as ⇒ 0 = (25)^2 + 2 (-1.6) s ⇒ s = 195.3 m. (iii) v^2 = u^2 + 2as ⇒ 0 = (25)^2 + 2 (-9.8) s ⇒ s = 31.9 m.
- v 2 = u^2 +2as ⇒ 502 = 0 + (2)(a)(1500) ⇒ a = 0.83 m s-^2
- v^2 = u^2 + 2as ⇒ 0 = (7)^2 + 2(-9.8) s / s = 2.5(0) m ⇒ max. height = 2.5 + 1.30 / 3. m
- s = ut + ½ at^2 -1.30 = 7t – ½ (9.8)t^2 t = 1.59 s
2002 (b) A particle, with initial speed u, moves in a straight line with constant acceleration. During the time interval from 0 to t, the particle travels a distance p. During the time interval from t to 2t, the particle travels a distance q. During the time interval from 2t to 3t, the particle travels a distance r. Show that 2q = p + r. Show that the particle travels a further distance 2r − q in the time interval from 3t to 4t.
Slight Variations 2011 (a) A particle is released from rest at A and falls vertically passing two points B and C.
It reaches B after t seconds and takes 2𝑡 7 seconds to fall from^ B^ to^ C , a distance of 2.45 m. Find the value of t.
2007 (a) A particle is projected vertically downwards from the top of a tower with speed u m/s. It takes the particle 4 seconds to reach the bottom of the tower. During the third second of its motion the particle travels 29.9 metres. Find (i) the value of u (ii) the height of the tower.
1995 (b) A juggler throws up six balls, one after the other at equal intervals of time t , each to a height of 3 m. The first ball returns to his hand t seconds after the sixth was thrown up and is immediately thrown to the same height, and so on continually. (Assume that each ball moves vertically). Find (i) the initial velocity of each ball. (ii) the time t. (iii)the heights of the other balls when any one reaches the juggler’s hand.
Collisions
1. At the point of collision s 1 = s 2 so get an equation for s 1 and s 2 (using s = ut + ½ at^2 ). Then equate the two equations and solve. 2. A final note on the distinction between distance and displacement (for vertical motion). On the way up, distance and displacement will be the same, but not when the ball is coming down (distance is total distance travelled, while displacement is only the height above the ground).
Note that in our equations of motion s always corresponds to displacement, not distance.
3. If you are asked for the distance that the ball travelled when collision occurred you will first have to establish whether the ball was on the way up or on the way down at that point. The easiest way to establish whether the ball was on the way up or the way down is to compare the time to reach maximum height to the time of collision, e.g. if the collision occurred after 5 seconds and the time to reach maximum height was 6 seconds the ball never reached the top and so was travelling up when collision occurred.
If the collision happened when the ball was on the way up then distance travelled equals s.
4. If the collision happened when the ball was on the way down then to find distance travelled you have to break the problem up into two parts; distance from point of projection to top of the trajectory plus distance from top of trajectory to the point of collision on the way down. 5. Concept of t and (t+2) Ball A is thrown up into the air and two seconds later ball B is thrown up. The balls collide after a further t seconds. Key: For the collision, A is in the air for t seconds and B is in the air for (t-2) seconds Or B is the air for t seconds and A is in the air for (t+2) seconds. The second option is preferable because it’s easier to deal with ‘pluses’ than ‘minuses’.
2004 (a) A ball is thrown vertically upwards with an initial velocity of 20 m/s. One second later, another ball is thrown vertically upwards from the same point with an initial velocity of u m/s. The balls collide after a further 2 seconds. (i) Show that u = 17.75. (ii) Find the distance travelled by each ball before the collision, giving your answers correct to the nearest metre.
1993 (b) A particle P is projected vertically upwards from the ground with an initial velocity of 47 m/s. Two seconds later another particle Q is projected vertically upwards from the same point with initial velocity 64.6 m/s. Calculate (i) how long Q is in motion before it collides with P. (ii) the height at which the collision occurs.
A particle falls freely under gravity from rest at a point p. After it has fallen for one second another particle is projected vertically downwards from p with a speed of 14.7 m/s. By considering the relative motion of the particles, or otherwise, find the time and distance from p at which they collide. Show the motion of both on a time-velocity graph.
2012 (a) A particle falls from rest from a point P. When it has fallen a distance 19·6 m a second particle is projected vertically downwards from P with initial velocity 39·2 m s−1. The particles collide at a distance d from P. Find the value of d.
1991 (b) A particle P is projected vertically upwards with an initial velocity u and two seconds later a second particle Q is projected vertically upwards from the same point with initial velocity 1.5 u. Calculate, in terms of u , how long Q is in motion before it collides with P and prove that | u | > 9.8.
2001 (b) A particle is projected vertically upwards with an initial velocity of u m/s and another particle is projected vertically upwards from the same point and with the same initial velocity T seconds later.
(i) Show that the particles will meet ( ) seconds from the instant of projection of the first particle
(ii) Show that the particles will meet at a height of metres.
T u g
u g T g
