Saturday, January 17, 2015

Physics 9702 Doubts | Help Page 46

  • Physics 9702 Doubts | Help Page 46



Question 275: [Kinematics > Linear motion]
An astronaut drops an apple on the surface of the Moon where the acceleration due to gravity is 1/6 of that on Earth. The time it takes for the apple to fall to the ‘ground’ compared with an apple dropped from the same height on Earth is
A the same
B √6 times as long
C 6 times as long
D 36 times as long

Reference: ???



Solution 275:
Answer: B.
EITHER
s = ut + ½ at2
Height = s. Initial velocity, u = 0. Let acceleration due to gravity on Earth = g. Acceleration due to gravity on Moon = g/6.

For apple falling on Earth,
s = 0 + 0.5gtE2
tE = √(2s / g)

For apple falling on Moon,
s = 0 + 0.5 (g/6) tM2
tM = √(12s / g)

Comparing the times on Moon and Earth,
tM / tE = [√(12s / g)] / [√(2s / g)] = √6 [√(2s / g)] / [√(2s / g)] = √6
tM / tE = √6
So, tM = √6tE

The time taken to fall on moon is √6 times as long as that on Earth.


OR
Let the acceleration due to gravity on Earth = g

Consider the apple falling on Earth.
v2 = u2 + 2as
Initial speed, u = 0. Final velocity, v = ??? Acceleration, a = g. Height = s
Final velocity reach by apple (at ground) = √[02 + 2gs] = √(2gs)

Acceleration, a = (v – u) / t
Time taken to reach ground on Earth , tE = (v – u) / a = (√(2gs) – 0 ) / g = √(2gs) / g

Consider the apple falling on Moon.
v2 = u2 + 2as
Initial speed, u = 0. Final velocity, v = ??? Acceleration, a = g / 6. Height = s
Final velocity reach by apple (at ground) = √[02 + 2(g/6)s] = √(gs / 3)

Acceleration, a = (v – u) / t
Time taken to reach ground on Moon, tM = (v – u) / a = (√(gs / 3) – 0 ) / (g/6) = 6√(gs) / g√3

Time taken to reach ground on Earth , tE = √(2gs) / g = √2 [√(gs) / g]
Time taken to reach ground on Moon, tM = 6√(gs) / g√3 = (6/√3) [√(gs) / g]
Let [√(gs) / g] = x

Comparing tM and tE,
tM / tE = [(6/√3) x] / [√2 x] = 6 / (√3 √2) = 6 / √6 = (√6 √6) / √6 = √6
tM / tE = √6
So, tM = √6tE

The time taken to fall on moon is √6 times as long as that on Earth.


The 1st method is obviously better and quicker than the 2nd one.








Question 276: [Dynamics > Momentum]
A small explosion is used to separate a space capsule from its rocket, after the rocket fuel is used up.

In this process, what happens to the total momentum and total kinetic energy of the system consisting capsule plus rocket?
            Total momentum                     Total kinetic energy
A         Increase                                               Increase
B         Increase                                               Unchanged
C         Unchanged                                          Unchanged
D         Unchanged                                          Increase

Reference: ???



Solution 276:
Answer: D.
At this level, systems are usually considered to be isolated.
For any isolated system, the total momentum is always conserved. So, the total momentum in the system is conserved. [A and B are incorrect]

The explosion releases energy. This is converted into kinetic energy of the rocket and capsule. So, kinetic energy increases.









Question 277: [Dynamics > Newton’s laws of motion]
The diagram below shows five wooden blocks joined by inelastic strings. A constant force accelerates the blocks to the right on a frictionless horizontal table.

In which string is the tension the greatest?
A W                B X                 C Y                 D Z

Reference: ???



Solution 277:
Answer: D.
To be able to solve this question we need to understand what’s happening in the system using the principles of physics.

Label the blocks, from left to right, as A, B, C, D and E respectively.
As the accelerating force is applied on block E, it (block E) moves to the right. All the other blocks also move to the right at the same speed as block E. So, there is a resultant force on each block and this resultant force is the same for all blocks.

Note that the ‘accelerating force’ acts on block E only. As block E moves to the right, there is a tension in string Z, connecting block E and block D. This tension is towards the left. (The direction of the tension is obvious. Imagine that the tension acted towards the right. The string would be loose. In the situation described in the question, the string can only drag the other block when it is straight.)

Assume that the blocks are all identical and let the mass of a block = m.

Consider the motion of block E alone, and the forces acting on it.
Let the accelerating force = F and let the tension in string Z = TZ.
Resultant force on block E (towards the right) = ma
Note that this resultant force is the same for each block, that is each block will accelerate with a resultant acceleration a.
Resultant force on block E = F – TZ = ma


Now, consider the tension in string Z, TZ. It acts to the left. From Newton’s third law, block D will exerts a force equal in magnitude and opposite in direction to tension in string Z. Therefore, there is a force towards the right on block D with its magnitude equal to TZ.

This force on block D will cause it to have a resultant acceleration ‘a’ towards the right (same as block E). But again, there will be a tension in string Y, connecting block C and block D. This tension would act towards the left. Let this tension be TY.

Consider the motion of block D alone, and the forces acting on it.
Resultant force on block D = ma
There is a force towards the right equal in magnitude of TZ and the tension in spring Y is to the left (as has been described above).
Resultant force on block D = TZ – TY = ma
Since the block moves towards the right, TZ should be greater than TY. TZ is greater than TY by an amount of ‘ma’.

In this way, the tensions in the other strings may be calculated and found to be smaller than TZ.
Tensions: TZ > TY > TX > TW









Question 278: [Kinematics > Linear motion > Graph]
A ball is released from rest above a horizontal surface. It bounces once and is caught.
(i) Which graph represents the variation with time t of the speed v of the ball?
(ii) Which graph represents the variation with time t of the velocity v of the ball?


Reference: Past Exam Paper – November 2012 Paper 13 Q9



Solution 278:
(i) Answer: A.
(ii) Answer: D.
For part (i), the question asks for the variation with time t of the speed v of the ball. Speed is a scalar and cannot have a negative velocity. (note that this is not general for all scalars. For example, temperature can have negative values even though they are scalars)

They ball is released from rest. So, initially both speed and velocity are zero. [B and C are incorrect for both cases]

The speed increases until it touches the surface. During collision with the surface, the speed decreases slightly in an instant. Then, the ball moves upwards but the speed decreases due to the acceleration due to gravity which acts downwards. When the ball is caught, its speed is zero. So, graph A represents the variation with time t of the speed v of the ball.

As for the velocity of the ball, it increases from zero as the ball falls (the downward velocity is taken as positive) until it reaches the surface where its velocity is maximum. After collision with the surface, the ball moves upwards. So, the velocity is now negative. The velocity of the ball moving upwards decreases (since the acceleration due to gravity is downwards) until it is caught (it’s no longer moving, so the velocity is zero). So, graph D represents the variation with time t of the velocity v of the ball.





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