Momentum,acceleration,force questions

Question

Take g-10m/s^-2

Give answers to 2 significant figures.

A racing bike is being put through its paces on a straight stretch of road by a stunt man.  The following velocity-time graph was obtained upon analysis of the bike’s performance characteristics:

http://www.flickr.com/photos/99973477@N03/11596100736/

1.        

a) Calculate the distance travelled by the bike during the first 40 seconds.

b) What was the acceleration of the bike after it started from rest?:

(i)        5seconds

(ii)       22seconds

(iii)      50seconds

2. On the Penetecost Islands there is an initiation rite called land diving. The islanders jump from a high platform with a rope made of vines attached to their ankles. The ropes are designed to bring the land divers to a stop just centimeters above the ground after they have fallen.

a)Calculate the velocity of a land diver assuming he starts from rest, falls 25meters, and that friction is negligible

b) How long does it take for the diver to fall the 25meters under the influence of gravity?

After travelling 25meters the vines attached to the land diver’s ankles start to stretch and cause the land diver to decelerate and come to a stop. The vine stretches by 50centimeters.

c)  Calculate the magnitude of the deceleration of the land diver whilst stopping.

3. A block on the end of a string 3meters long is swung in an anticlockwise horizontal circle with a constant speed of 5m/s. You are looking down (bird’s-eye view) on to this horizontal circle.

http://www.flickr.com/photos/99973477@N03/11595743473/in/photostream/

a)  On the diagram above draw, using arrows, the instantaneous velocity of the block at the South (S) position and at the East (E) position.

b)  What is the change in velocity (magnitude and direction) of the block as its motion changes from the south (S) position to the East (E) position in its circular path?

4. A car of mass 800kg accelerates from rest to 20m/s^-1 in 8.0s. The resistance forces acting on the car total 1000N.

a)Draw a well labelled diagram showing all forces acting on the car. Clearly show the point of application of each force, ie, show where each force starts.

b)Calculate the driving force being provided by the engine of the car.

5.  http://www.flickr.com/photos/99973477@N03/11595541175/in/photostream/

a) On the diagram above, draw and label two arrows that show the action-reaction forces acting on the rope between the boat and the skier. Use the letters ‘A’ for action, and ‘R’ for reaction next to your arrows.

b) Complete the two sentences below which describe the action-reaction pair above.

Action: The force of the..........on the............

Reaction: The force of the  ............on the...........

c) Calculate the magnitude (size) of the drag (resistance) forces acting on the skier.

6. A toy car of mass 50grams travels down a smooth incline at 30 degrees to the horizontal. Friction may be ignored.

a) Calculate the net force acting on the car as it rolls down the slope.

b) Calculate the force of the incline on the car as it travels down the slope.

Answer 1

1)

a) Distance traveled by the bike during the first 40 seconds.

distance = velocity * time

= 40*40 = 1600 m.

 

b) Acceleration = Change in velocity/Change in time.

(i)

Observe the graph :

If x changes from 0 to 5 sec, then y changes from 0 to 15 m/s.

Acceleration = Change in velocity/time.

                  = (15-0)/(5-0)

                 =15/5

                =3 m/s²

(ii)

Observe the graph :

If x changes from 0 to 22 sec, then y changes from 0 to 60 m/s.

Acceleration = Change in velocity/time.

                  = (60-0)/(22-0)

                  =60/22

                 =2.73 m/s²

 

(iii)

Observe the graph :

If x changes from 0 to 50 sec, then y changes from 0 to 20 m/s.

Acceleration = Change in velocity/time.

                  = (20-0)/(50-0)

                  =20/50

                 =0.4 m/s² (deceleration)  

(2).

Initial velocity ,u =0

Falling distance ,s=25 m

Acceleration, a= g =10 m/s^2

a)

v^2-u^2 =2as

v^2-u^2 =2gs

v^2=2(10)(25)

v^2=500

v=10√5 m/s

b)

s=ut+(1/2)at^2

25=0+(1/2)(10)t^2

t^2=5

t=√5 s

c)

s= 25+0.5 =25.5

v=0

u=10√5 m/s

v^2-u^2 =2as

0-(10√5)^2=2(a)(25.5)

-500=51a

a=-9.804 m/s^2.