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Practice each question, then open the answer or worked solution.
Give an example of a situation in which an applied force does not result in a change in kinetic energy.
Work done during the circular motion there is no change in the kinetic energy.
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of the same magnitude. How would the distance moved by them before coming to rest compare?
KE1 = KE2
WD1 = WD2
F1s1 = F2s2
F1 = F2
s1 = s2
A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in the figure. What will be the trajectory of the particle if the string is cut at
a) point B
b) point C
c) point X
a) The trajectory of the particle when the string is cut at point B is the tangential velocity will be vertically downward and the bob will move along the same path.
b) The trajectory of the particle when the string is cut at point C is the tangential velocity will be horizontal and will move in half parabolic path.
c) The trajectory of the particle when the string is cut at point X is when angle θ is formed and will move in the parabolic path till the bob reaches the highest point.
Short AnswerA graph of potential energy V(x) versus x is shown in the figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.
From the given graph of KE versus x
| x | KE | Point |
| 0 | 0 | A |
| 0 | E1 | B |
| X1 | E0 | C |
| X2 | E0 | D |
| X3 | 0 | F |
From the below graph of velocity versus x
| Point | V | X |
| A | 0 | 0 |
| B | ±V1 | 0 |
| C | ±V0 | X1 |
| D | ±V0 | X2 |
| E | 0 | X3 |
A ball of mass m, moving with speed 2v0, collides inelastically with an identical ball at rest. Show that
a) for a head-on collision, both the balls move forward
b) for a general collision, the angle between the two velocities of scattered balls is less than 90o.
a) Let v1 and v2 be the velocities of the two balls after the collision. According to the law of conservation of momentum,
mv0 = mv1 + mv2
v2 = v1 + 2ev0
e < 1
b)
According to the law of conservation of momentum,
p = p1 + p2
When the angle between the p1 and p2 is 90o, p > p1 + p2
Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C, and D in which the relation between potential energy V, kinetic energy (K) and total energy is as given below:
Region A: V > E
Region B: V < E
Region C: K > E
Region D: V > K
State with reason in each case whether a particle can be found in the given region or not.
For region A, E = V + K and V > E which means that the KE is negative and this is not possible.
For region B, K = E – V and V < E which means that both the energies are greater than zero
For region C, V = E – K and K > E which means that PE is negative
For region D, K = E – V and V > K which means that PE is greater than KE
The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in the figure. If the length of the pendulum is 1 m, calculate
a) the height to which bob A will rise after collision
b) the speed with which bob B starts moving.
Neglect the size of the bobs and assume the collision to be elastic.
a) There is not much rise in bob A after the collision as the PE of bob A gets converted in KE and the momentum is transferred to bob B.
b)
The speed of bob B is found using the sum of KE of bob A and KE of bob B = 4.42 m/s
A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m/s. Calculate
a) the loss of PE of the drop
b) the gain in KE of the drop
c) is the gain in KE equal to the loss of PE. If not, why?
a) PE at the highest point = 10 J
b) Gain in KE = 1/2 mv2 = 1.250 J
c) Gain in KE is not equal to the PE
Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in common. One of the bobs is released after being displaced by 10o so that it collides elastically head-on with the other bob.
a) describe the motion of two bobs
b) draw a graph showing variation in energy of either pendulum with time for 0 ≤ t ≤ 2T where T is the period of each pendulum
| Time t | KEa | PEa | Ea | KEb | PEb | Eb |
| 0 | 0 | 0 | 0 | 0 | E | E |
| T/4 | E | 0 | E | 0 | 0 | 0 |
| 2T/4 | E | 0 | 0 | 0 | E | E |
| 3T/4 | E | 0 | E | 0 | 0 | 0 |
| 4T/4 = T | 0 | 0 | 0 | 0 | E | E |
Suppose the average mass of raindrops is 3.0 × 10-5 kg and their average terminal velocity 9 m/s. Calculate the energy transferred by rain to each square meter of the surface at a place which receives 100 cm of rain in a year.
Energy transferred by the rain to the surface of the earth = 1/2 mv2
The velocity of the rain = 9 m/s
Mass = (volume)(density) = 1000 kg
Energy transferred by 100 cm rainfall = 1/2 mv2 = 4.05×104 J