Thursday, 30 December 2010

Mechanics1 Worked solutions June 2006 paper

Link to worked solutions for June 2006 M1 paper:

Notes to support worked solutions:

1. (a) the gradient of a speed-time graph is acceleration so a straight line (so constant gradient) will represent constant acceleration
(b) so a horizontal line has gradient = 0 so acceleration = 0 so constant speed
(c) area under a speed time graph represents distance travelled

2. (a) considering the whole system the inital momentum will be equal to the final momentum - remember momentum = mass x velocity - make sure directions of velocities are used correctly - in this case right is defined as positive - the only unknown v is found - the fact that it is found to b positive shows that the direction is to the right so the direction of the particle is unchanged
(b) considering just a single particle the change in momentum (final momentum - initial momentum) is equal to the impulse acting on it

3.(a) for the journey A to B, s&u&t are known so a can be found
(b) for the whole journey A to C, s&u are known and a was found in (a) so v can be found
(c) again using whole journey A to C the only remaining unknown t can be found - don't forget to subtract 2 seconds to give the required answer of the time from B to C

4. aside about the angle of elevation - tan of the angle is given as 3/4 which could be used to find the angle although this would be a long decimal - it is simpler for the question to find that sin of the angle = 3/5 and cos of the angle = 4/5 gby using a 3,4,5 triangle
(a) perpendicular to the plane finds the normal reaction
parallel to the plane finds the frictional force
these are combined to find the coefficient of friction
(b) frictional force is now up the slope althought its size will not have changed from (a) because the normal reaction won't have changed because no forces perpendicular to the slope have changed
- find the resultant force down the slope (ie parallel to the slope)
- use F=ma to find the acceleration down the slope

5. (a) the information given about the tensions and just using up=down lets us find the size of the tensions
(b) the diagram was set up with x being half the length AB - by taking moments about this length can be found
(c) taking moments about B eliminates the unknown W so the tension can be found then up = down finds W

6. (a) considering whole system (tensions cancel each other out) and using resultant force = mass x acceleration the acceleration is found
(b) by considering just the car the tension can be found (could have equally considered just the trailer
(c) as diagram shows the only forces acting on car are the engine and the resistance - again use F = ma to find the acceleration - followed by suvat to find the distance

7. (a) speed is the magnitude of the velocity vector so just use pythagoras on the i and j values
(b) make sure the correct angle is identified for the bearing (clockwise from north)
(c) the position after 3 hours is the start position plus 3 lots of the velocity
(d) position after 2 hours is the start position plus 2 lots of the velocity
- the new velocity just adds 5 to the j every hour
(e) to be due east of a point means having the same vertical distance (ie the j component as that point) - be careful translating 1.2 hours into 1 hour and 12 minutes (12 mins is 2/10 of an hour)
(f) 2 hours after the new velocity just adds 10 the the j
the displacement vector from R is found by subtracting R
then the distance is the magnitude of the displacement vector so just use pythagoras on the i and j components

Tuesday, 21 December 2010

Mechanics1 Worked solutions Jan 2006 paper

The link is to the worked solutions for the January 2006 Mechanics 1 paper and then there are some notes below explaining the solutions.

1. The particle is going to travel up to a maximum height and then falll to the ground. Known suvat variables for the whole journey are u,a,t so h and v can be found to answer the the two parts of the question.

2. (a) Remember momentum = mass x velocity and remember initial momemtum = final momentum if whole system considered. Only unknown is the veolicty to be found.
(b) (i) Again only unknown considering momentum of whole system is the mass of Q to be found.
(ii) Remember impulse = change in momentum. Interested in impulse on Q so consider particle Q only and find change in momentum. Remember the velocities are in opposite directions so must have opposite signs.

3. (a) Considering beam as uniform so weight of beam through pivot. By taking moments about the pivot the weight of the beam and the reaction at the beam are not involved.
(b) Uniform means weight acts at midpoint
(c) Now weight (15g) of beam moves away from centre. It doesn't matter which side it is put as a negative solution will mean you put it thw wrong side. Again moments about pivot eliminates reaction at pivot.

4. Resultant of P and Q is R. So diagrams shows P followed by Q being the same as R. If needed the cosine rule is given in formula book under Core2. Make sure solutions are sensible.

5. Best to work perpendicular and parallel to the plane so that one of the unknowns can be eliminated (F when working perpendicular). So P and the weight need to be split into two components.
(a) perpendicular to the plane P is only unknown
(b) parallel to the plane - now P is known F is only unknown - and as R is known, the coefficient of friction is known (remember max friction = coefficient of friction x normal reaction)
(c) best to redraw diagram with P removed. Remember friction will now be up the slope as it is on point of moving down slope. Normal reaction will have changed so recalculated. Coefficient of friction is a function of the particle and slope so doesn't change (use value calculated in (b)). The maximum friction force is greater than the force down the slope so the particle doesn't move.

6. (a) speed is magnitude of velocity
(b) remember to find angle and then form into a bearing
(c) to find position after time t is initial position + t x velocity
for the A and B to collide the i and j parts of their position must both match at the same time (they do at t=7)
then substitute t=7 into either position (should give same solution)

7. Remember tension will be the same on both sides. Larger particle will move down the slope and smaller particle up.
(a) considering just particle A (smooth slope)
only force down is component of weight
only force up is tension
using Newtons second law
resultant force down slope = mass x acceleration
(b) now particle B (rough slope)
perpendicular to find expression for normal reaction
particle is moving up
only force up is tension (known from (a))
forces down are component of weight and friction (reaction known so coefficient of friction is only unknown)
resultant force up slope = mass x acceleration
(c) forces on pulley are vertical components of tension

Monday, 15 November 2010

Mechanics1 Moments 15/11/10

  • Turning effect (Moments)
  • Combining moments
  • Moments in systems in equiibrium
  • Problems involving moments
  • Problems involving multiple pivots
  • Problems when about to tilt about one pivot

Wednesday, 10 November 2010

Mechanics1 Collisions, guns and jerks 10/11/10

  • Using the principle of conservation of momentum in different situations
  • Colliding particles
  • Guns firing bullets
  • Jerks on a string

Wednesday, 20 October 2010

Mechanics1 Momentum and impulse 20/10/10

  • Momentum = mv
  • Change in momentum
  • Impulse = Ft = Change in momentum
  • Problems involving impulse & change in momentum
  • Principle of conservation of momentum
  • Problems involving collision or particles

Wednesday, 13 October 2010

Mechanics1 Connected particles 13/10/10

  • Connected particles - equal and opposite tension forces in a string
  • Treating each particle separately
  • Using F=ma where acceleration is common to both particles
  • Problems involving connected particles

Monday, 11 October 2010

Mechanics1 Newton's laws 11/10/10

  • Definitions of Newton's laws
  • Using F=ma
  • Problems involving use of F=ma and equations of motion

Wednesday, 6 October 2010

Monday, 4 October 2010

Mechanics1 Types of forces 04/10/10

  • Force vectors
  • Equilibrium
  • Weight, normal reaction force, tension, thrust, friction
  • Problems involving systems of forces in equilibrium
  • Using the directions parallel and perpendicular to a slope

Monday, 27 September 2010

Mechanics1 Resolving forces 27/09/10

  • Resultant of a horizontal and vertical force
  • Resolving a force into horizontal and vertical components
  • Resultant of two or more forces
  • Using vectors for forces

Mechanics1 Vertical motion 27/09/10

  • Recap motion in horizontal plane
  • Vertical motion - acceleration due to gravity
  • Example problem with vertical motion

Wednesday, 22 September 2010

Mechanics1 Equations of motion 22/09/10

  • Introduce equations of motion from first principles
  • Summary of equations of motion
  • Problems involving equations of motion in a horiztonal plane

Monday, 20 September 2010

Mechanics1 Applications of vectors 20/09/10

  • Reminder of finding magnitude and direction of a vector
  • Find the i-j components of a vector given magnitude and direction
  • Finding a unit vector in the direction of a given vector
  • Defining displacement, position, velocity and acceleration vectors
  • Solving problems involving vectors

Friday, 17 September 2010

Monday, 13 September 2010

Mechanics1 Bearings and vectors 13/09/10

  • Problems involving bearings - using cosine rule and sine rule
  • Recapping basics of vectors
  • Vectors in i-jform
  • Combining vectors
  • Parallel vectors