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# Mechanics 1 - Kinematics

A revision aid for Kinematics (Edexcel Mechancis 1)

by

Tweet## Lydia Blain

on 11 March 2013#### Transcript of Mechanics 1 - Kinematics

June 2008, Q2 At time t = 0, a particle is projected vertically upwards with speed u m/s from a point 10 m above the ground. At time T seconds, the particle hits the ground with speed 17.5 m/s. Find

(a) the value of u, (3)

(b) the value of T. (4) June 2001, Q3 Edexcel Mechanics 1

revision resource Kinematics Kinematics Edexcel Mechanics 1

revision tool for chapter 2. Kinematics is the study of motion of objects or particles; their displacement, speed and velocity. It does not consider the causes of the motion (forces). Work through the questions and tutorial resources provided. At the end of the revision process you should be ready to try some past exam questions. You should allocate a minimum of 2 hours for your Kinematics revision. June 2007, Q4 Examiner top tip:

Although you may use more accurate values for g in other subjects (like 9.81) using anything other than 9.8 in Mechanics 1 will be penalised.

Because g is accurate to 2 s.f. your answer should also be given to 2 (or 3) s.f. Anything more (or less) accurate will also be penalised. A car of mass 1200 kg moves along a straight horizontal road. In order to obey a speed restriction, the brakes of the car are applied for 3 s, reducing the car’s speed from 30 m s1 to 17 m s1. The brakes are then released and the car continues at a constant speed of 17 m s1 for a further 4 s. Figure 2 shows a sketch of the speed-time graph of the car during the 7 s interval. The graph consists of two straight line segments.

(a) Find the total distance moved by the car during this 7 s interval. (4)

(b) Explain briefly how the speed-time graph shows that, when the brakes are applied,

the car experiences a constant retarding force. (2)

(c) Find the magnitude of this retarding force. (3) A car is moving along a straight horizontal road. At time t = 0, the car passes a point A with speed 25 m s–1. The car moves with constant speed 25 m s–1 until t = 10 s. The car then decelerates uniformly for 8 s. At time t = 18 s, the speed of the car is V m s–1 and this speed is maintained until the car reaches the point B at time t = 30 s.

(a) Sketch a speed–time graph to show the motion of the car

from A to B. (3)

Given that AB = 526 m, find

(b) the value of V, (5)

(c) the deceleration of the car between t = 10 s and t = 18 s. (3) Examiner top tip:

When asked to find the deceleration, you should expect to calculate a negative answer for 'a' but you should then quote deceleration as a positive value.

e.g. a=-3, so deceleration=3.

Similar rules apply when asked for a value for speed, when you have calculated a velocity. Spot the mistake! Look at the exam solution below. See if you can spot the mistake. Then see if you can find the correct answer. u=10.5 (m/s)

T=2.86 (s) Did you get the correct answer?

If not, watch the video solution. (a) Total distance = area under curve

Distance = 0.5(30+17)x3 + 4x17

= 138.5 (m)

M1 – attempt at area of trapezium or triangle and rectangle

A1 correct equation

M1 – equations for total area under curve

A1 correct answer only (cao)

(b) Straight line graph represents constant deceleration

F=ma means the Force is constant

M1 comment about straight line or gradient representing acceleration (or deceleration)

A1 cao

(c) Deceleration = (30-17)/3

F= 1200 x (30-17)/3 = 5200N

M1 – correct statement about change of velocity over time

M1 - correct use of F=ma

A1 - cao Mark scheme and notes (b) v=11 (m/s)

(c) a=1.75 Did you get the correct answer?

If not, watch the video solutions. Examiner top tip:

When considering vertical motion assign a direction as positive. If upward is positive, then a=-9.8. If downwards is positive then a=+9.8. Even if an object changes direction during the period of motion, reference everything to the initial state. For example, if the object was initially thrown upwards, with up assigned as positive, but lands lower down, then u = +, v = -, s = - Did you spot it? Part (c) was incorrect.

The time is not 2s at B.

Instead use s=72 (m) to find that v=43.3 (m/s)

Full transcript(a) the value of u, (3)

(b) the value of T. (4) June 2001, Q3 Edexcel Mechanics 1

revision resource Kinematics Kinematics Edexcel Mechanics 1

revision tool for chapter 2. Kinematics is the study of motion of objects or particles; their displacement, speed and velocity. It does not consider the causes of the motion (forces). Work through the questions and tutorial resources provided. At the end of the revision process you should be ready to try some past exam questions. You should allocate a minimum of 2 hours for your Kinematics revision. June 2007, Q4 Examiner top tip:

Although you may use more accurate values for g in other subjects (like 9.81) using anything other than 9.8 in Mechanics 1 will be penalised.

Because g is accurate to 2 s.f. your answer should also be given to 2 (or 3) s.f. Anything more (or less) accurate will also be penalised. A car of mass 1200 kg moves along a straight horizontal road. In order to obey a speed restriction, the brakes of the car are applied for 3 s, reducing the car’s speed from 30 m s1 to 17 m s1. The brakes are then released and the car continues at a constant speed of 17 m s1 for a further 4 s. Figure 2 shows a sketch of the speed-time graph of the car during the 7 s interval. The graph consists of two straight line segments.

(a) Find the total distance moved by the car during this 7 s interval. (4)

(b) Explain briefly how the speed-time graph shows that, when the brakes are applied,

the car experiences a constant retarding force. (2)

(c) Find the magnitude of this retarding force. (3) A car is moving along a straight horizontal road. At time t = 0, the car passes a point A with speed 25 m s–1. The car moves with constant speed 25 m s–1 until t = 10 s. The car then decelerates uniformly for 8 s. At time t = 18 s, the speed of the car is V m s–1 and this speed is maintained until the car reaches the point B at time t = 30 s.

(a) Sketch a speed–time graph to show the motion of the car

from A to B. (3)

Given that AB = 526 m, find

(b) the value of V, (5)

(c) the deceleration of the car between t = 10 s and t = 18 s. (3) Examiner top tip:

When asked to find the deceleration, you should expect to calculate a negative answer for 'a' but you should then quote deceleration as a positive value.

e.g. a=-3, so deceleration=3.

Similar rules apply when asked for a value for speed, when you have calculated a velocity. Spot the mistake! Look at the exam solution below. See if you can spot the mistake. Then see if you can find the correct answer. u=10.5 (m/s)

T=2.86 (s) Did you get the correct answer?

If not, watch the video solution. (a) Total distance = area under curve

Distance = 0.5(30+17)x3 + 4x17

= 138.5 (m)

M1 – attempt at area of trapezium or triangle and rectangle

A1 correct equation

M1 – equations for total area under curve

A1 correct answer only (cao)

(b) Straight line graph represents constant deceleration

F=ma means the Force is constant

M1 comment about straight line or gradient representing acceleration (or deceleration)

A1 cao

(c) Deceleration = (30-17)/3

F= 1200 x (30-17)/3 = 5200N

M1 – correct statement about change of velocity over time

M1 - correct use of F=ma

A1 - cao Mark scheme and notes (b) v=11 (m/s)

(c) a=1.75 Did you get the correct answer?

If not, watch the video solutions. Examiner top tip:

When considering vertical motion assign a direction as positive. If upward is positive, then a=-9.8. If downwards is positive then a=+9.8. Even if an object changes direction during the period of motion, reference everything to the initial state. For example, if the object was initially thrown upwards, with up assigned as positive, but lands lower down, then u = +, v = -, s = - Did you spot it? Part (c) was incorrect.

The time is not 2s at B.

Instead use s=72 (m) to find that v=43.3 (m/s)