Tuesday, December 2, 2014

Physics 9702 Doubts | Help Page 31

  • Physics 9702 Doubts | Help Page 31

Question 193: [Dynamics > Momentum]
Two train carriages each of mass 5000 kg roll toward one another on level track. One is travelling at 2.00 m s–1 and other at 1.00 m s–1, as shown.

They collide and join together.
What is the kinetic energy lost during collision?
A 1250 J                      B 7500 J                      C 11 250 J                   D 12 500 J

Reference: Past Exam Paper – June 2014 Paper 12 Q7

Solution 193:
Answer: C.
Total Kinetic Energy before collision = ½ (5000)(2)2 + ½ (5000)(1)2 = 12500J

From the conservation of momentum, the sum of momentum before collision is equal to the sum of momentum after collision.

Let the final speed after collision = v
Take the direction towards the right as the positive direction.

5000(2) + 5000(-1) = 10000v
Speed, v = 5000/10000 = 0.5ms-1

Kinetic Energy after collision = ½ (10000)(0.5)2 = 1250J

Kinetic Energy lost = 12500 – 1250 = 11 250J

Question 194: [Kinematics > Linear motion > Graph]
In an experiment to determine acceleration of free fall using a falling body, what would lead to value that is too large?
A air resistance
B dimensions of body are too large
C measured distance longer than true distance
D measured time longer than true time

Reference: Past Exam Paper – June 2012 Paper 11 Q6

Solution 194:
Answer: C.
The measurements that can be taken for a falling body are the distance between 2 points, and the time that the falling body takes to travel between the 2 points. From these measurements, a graph of distance against (time)2 may be plotted to obtain the acceleration of free fall, for example.

Air resistance acts against motion, and thus causes a resultant acceleration on the body which is less than the acceleration of free fall.

Dimension of the body would tell about the size (e.g. volume, …) of the body. This information is not necessary for the determination of the acceleration of free fall.

As explained above, the acceleration may be obtained by the gradient of a distance – (time)2 graph. Gradient = distance / (time)2. So, having a longer measured distance than the true distance causes the gradient to is larger, while having a longer measured time than the true time causes the gradient to be smaller.

Question 195: [Kinematics & Dynamics]
Bullet of mass 2.0 g is fired horizontally into block of wood of mass 600 g. Block is suspended from strings so that it is free to move in vertical plane. Bullet buries itself in block. Block and bullet rise together through vertical distance of 8.6 cm, as shown.

(i) Calculate change in gravitational potential energy, ΔEp of block and bullet
(ii) Show that initial speed of block and bullet, after they began to move off together, was 1.3ms–1

(b) Using information in (a)(ii) and principle of conservation of momentum, determine speed of bullet before impact with block

(i) Calculate kinetic energy of bullet just before impact
(ii) Explain what can be deduced from answers to (c)(i) and (a)(i) about type of collision between bullet and block

Reference: Past Exam Paper – June 2005 Paper 2 Q3

Solution 195:
{Total mass = mass of bullet + mass of block = 2 + 600 = 602g = 0.602g}
ΔEp = mgΔh = 0.602 x 9.8 x 0.086 = 0.51J

{v2 = u2 + 2as = 0 + 2gh. For the OR case, KE of bullet and block = Change in gravitational PE. ½ mv2 = 0.51J}
v2 (= 2gh) = 2 (9.8) (0.086)                 or v2 = (2 x 0.51) / 0.602
Initial speed of block and bullet, v = 1.3ms-1 

{Sum of momentum before collision = Sum of momentum after collision}
2v = 602 x 1.3
Speed of bullet, v = 390ms-1

Kinetic energy of bullet, Ek = ½mv2 = ½ (0.002) (3902) = 152J or 153J or 150J

The kinetic energy is not the same / changes OR the kinetic energy before the impact is greater than the kinetic energy / potential energy after the impact. So, the collision must be inelastic.

Question 196: [Forces]
Car of mass m travels at constant speed up a slope at an angle θ to horizontal, as shown in the diagram. Air resistance and friction provide resistive force F.

What force is needed to propel car at this constant speed?
A mg cos θ
B mg sin θ
C mg cos θ + F
D mg sin θ + F

Reference: Past Exam Paper – June 2012 Paper 11 Q14

Solution 196:
Answer: D.
The car has a component of its weight acting (down) along the slope (in the same direction as the resistive force F). This component is equal to mg sinθ.

For the car to move at constant speed, the force required to move the car (up) along the slope should be equal to the sum of the forces acting (down) along the slope such that the resultant force and hence resultant acceleration is zero.

Thus, the force = mg sinθ + F


  1. in solution 193 how is 5000(2)+5000(-1)=10000???

    1. its 10000v

      we are equating momentum. it's an equation

  2. can someone explain question 195"b"

  3. it seems to be self-explicit. What is your confusion?

  4. Replies
    1. just evaluate the left hand side of the equation above it, then make v subject of formula


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