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Laws of motion and system of particles (check your understanding)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Objective questions, contained in this module with hidden solutions, help improve understanding of the topics covered under the module "Laws of motion and system of particles".

The questions have been selected to enhance understanding of the topics covered in the module titled " Laws of motion and system of particles ". All questions are multiple choice questions with one or more correct answers. There are two sets of the questions. The “understanding level” questions seek to unravel the fundamental concepts involved, whereas “application level” are relatively difficult, which may interlink concepts from other topics.

Each of the questions is provided with solution. However, it is recommended that solutions may be seen only when your answers do not match with the ones given at the end of this module.

Understanding level (Laws of motion and system of particles)

Exercise 1

A spherical object placed on a smooth horizontal table is at rest. It suddenly breaks in two unequal parts as a result of chemical reaction taking place inside the object. Then, the parts of the object move :

(a) in same direction

(b) in different directions

(c) in opposite directions with equal speeds

(d) in opposite directions with unequal speeds

Solution

The particle is at rest before breaking into parts due to internal force. It means that the COM of the parts should remain at the initial position. In other words, its velocity should be zero even after breaking. Now, the speed of the COM is :

v COM = m 1 v 1 + m 2 v 2 m 1 + m 2 = 0 v COM = m 1 v 1 + m 2 v 2 m 1 + m 2 = 0

m 1 v 1 + m 2 v 2 = 0 m 1 v 1 + m 2 v 2 = 0

v 1 = - ( m 2 m 1 ) v 2 v 1 = - ( m 2 m 1 ) v 2

Clearly, the velocities of the parts of the sphere should move in opposite directions. As the parts are unequal, they should move with different speeds as well.

Hence, option (d) is correct.

Exercise 2

In a system of two particles of 0.1 kg and 0.2 kg, The particle of 0.1 kg is at rest, whereas other particle moves with an acceleration "a". The magnitude of acceleration of the COM of the system of two particles is :

(a) a 3 (b) 2 a 3 (c) 3 a 2 (d) a 3 (a) a 3 (b) 2 a 3 (c) 3 a 2 (d) a 3

Solution

The acceleration of the system of two particles is given by :

a COM = m 1 a 1 + m 2 a 2 m 1 + m 2 a COM = 0.1 x 0 + 0.2 x a 0.3 = 2 a 3 a COM = m 1 a 1 + m 2 a 2 m 1 + m 2 a COM = 0.1 x 0 + 0.2 x a 0.3 = 2 a 3

Hence, option (b) is correct.

Exercise 3

A sphere of radius "r" with uniform density is placed on a smooth horizontal surface. A horizontal force "F" is applied on the sphere. The acceleration of the center of sphere :

(a) is maximum when force is applied near the bottom edge of the sphere.

(b) is maximum when force is applied near the top edge of the sphere.

(c) is maximum when force is applied at middle of the sphere.

(d) is independent of position of application.

Solution

Sphere is composed by a system of particle. As the density of sphere is uniform, its COM coincides with the center of sphere. Now, the acceleration of the "center of mass" is dependent on the net external force on the system. It does not depend on the point of application. Recall that we explained that COM represents a point where external force "appears" to apply - not the point where net external force acts.

Hence, option (d) is correct.

Exercise 4

Two blocks of masses "m" and "2m" are released from rest as shown in the figure. The displacement of the COM of the block system (in meters) after 1 second is (neglect friction and masses of string and pulley) :

Figure 1
Blocks and pulley system
 Blocks and pulley system  (lmsq1.gif)

(a) 0.51 (b) 0.61 (c) 0.71 (d) 0.81 (a) 0.51 (b) 0.61 (c) 0.71 (d) 0.81

Solution

In the beginning when the blocks are released from rest, the COM of the two blocks system is also at rest. Once the blocks are released, they are acted upon by force due to gravity. We consider tensions in the string as the internal force. It means that COM is accelerated downward. In order to find the displacement of the COM, we need to know its vertical acceleration. The acceleration of the COM is obtained by :

a COM = m 1 a 1 + m 2 a 2 m 1 + m 2 a COM = m 1 a 1 + m 2 a 2 m 1 + m 2

a COM = a 1 + 2 a 2 3 a COM = a 1 + 2 a 2 3

Thus, we need to know the accelerations of individual block. As the blocks are connected with a taut string, the magnitude of acceleration of each block is same, but they are directed opposite to each other. From force analysis on individual block,

Figure 2
Free body diagram
 Free body diagram  (lmsq2.gif)

For first block :

F y = T - m g = m a F y = T - m g = m a

For second block :

F y = 2 m g - T = 2 m a F y = 2 m g - T = 2 m a

Solving two equations, we have :

a = g 3 a = g 3

Putting in the equation of acceleration of COM in y - direction with appropriate sign (downward direction as positive):

a COM = a 1 + 2 a 2 3 = - g 3 + 2 g 3 3 a COM = a 1 + 2 a 2 3 = - g 3 + 2 g 3 3

a COM = g 9 = 10 9 a COM = g 9 = 10 9

Now, let "y" be the displacement of COM in 1 second,

y = 1 2 a t 2 y = 1 2 a t 2

y = 1 2 x 10 9 x 1 = 0.51 m y = 1 2 x 10 9 x 1 = 0.51 m

Hence, option (a) is correct.

Exercise 5

A block of mass "m" is released from the top of an incline of mass "M" as shown in the figure, where all the surfaces in contact are frictionless. The displacement of the incline, when the block has reached bottom is :

Figure 3
Block and incline system
 Block and incline system  (lmsq3.gif)

(a) m L m + M (b) m M L m + M (c) m L M - m (d) m M L M - m (a) m L m + M (b) m M L m + M (c) m L M - m (d) m M L M - m

Solution

This question makes use of the concept of COM to a great effect. Answer to this question involving force analysis would have been difficult. Here, we note that the block and incline is at rest in the beginning. This means that COM of the system is at rest before the block is released. When the block is released, the system is subjected to external force due to gravity. This means that COM should move along the direction of applied external force. As the direction of external force is vertically downward, the COM will also be accelerated downward. Importantly, COM will have no displacement in horizontal direction. We shall make use of this fact in solving this problem.

We can estimate here that the block slides down towards left. In order that the COM has no horizontal displacement, we can conclude that incline moves towards right. Let us consider that the incline moves to right by a distance "x".

Figure 4
Block and incline system
 Block and incline system  (lmsq4.gif)

The horizintal component of displacement of block on reaching bottom is given as "L" with respect to incline. This component of displacement with respect to ground,"d", is (note that in the meantime incline has also moved in opposite direction and the block is carried along with the incline) :

d = L - x d = L - x

As COM of the "block and incline" system has no displacement in horizontal direction,

x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 = 0 x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 = 0

m 1 x 1 + m 2 x 2 = 0 m ( L - x ) - M x = 0 m 1 x 1 + m 2 x 2 = 0 m ( L - x ) - M x = 0

x = m L m + M x = m L m + M

Hence, option (a) is correct.

Application level (Laws of motion and system of particles)

Exercise 6

Two particles are projected with speeds v 1 v 1 and v 2 v 2 making angles θ 1 θ 1 and θ 2 θ 2 with horizontal at the same instant as shown in the figure. Neglecting air resistance, the trajectory of the COM of the two particles :

Figure 5
Two projectiles
 Two projectiles  (lmsq7.gif)

(a) can be a vertical straight line

(b) can be a horizontal straight line

(c) is a straight line

(d) is a parabola

Solution

To answer this question, we need to analyze the speeds of COM of two particles in horizontal (x) and vertical (y) directions. Here,

v COMx = m 1 v 1x + m 2 v 2x m 1 + m 2 = v COMx = v 1x + v 2x 2 v COMx = m 1 v 1x + m 2 v 2x m 1 + m 2 = v COMx = v 1x + v 2x 2

v COMx = v 1 cos θ 1 - v 2 cos θ 2 2 = a constant v COMx = v 1 cos θ 1 - v 2 cos θ 2 2 = a constant

and

v COMy = m 1 v 1y + m 2 v 2y m 1 + m 2 = v COMy = v 1y + v 2y 2 v COMy = m 1 v 1y + m 2 v 2y m 1 + m 2 = v COMy = v 1y + v 2y 2

v COMy = v 1 cos θ 1 - g t + v 2 cos θ 2 - g t 2 v COMy = v 1 cos θ 1 - g t + v 2 cos θ 2 - g t 2

v COMy = v 1 cos θ 1 + v 2 cos θ 2 - 2 g t 2 v COMy = v 1 cos θ 1 + v 2 cos θ 2 - 2 g t 2

On inspection of the speed of COM in x-direction, we see that it can be zero if :

v 1 cos θ 1 = v 2 cos θ 2 v 1 cos θ 1 = v 2 cos θ 2

For this condition, COM has only speed in vertical direction and varies with time. For all other values of attributes of motion, which do not satisfy the equality, the COM has both components of velocity and, therefore, moves in parabolic trajectory.

Note:
This is a case of two particles system, which is acted by the force due to gravity. If the initial directions of velocities and that of force are same, then COM would move in the direction of force. Here, situation is different, initial velocities and the force have different directions to begin with. As such the motion of COM is parabolic except for a condition as derived above.

Hence, option (a) is correct.

Exercise 7

The wedge "C" is placed on a smooth horizontal plane. Two blocks "A" and "B" connected with "mass-less" string is released over a double sided wedge "C". The masses of blocks "A","B" and "C" are m, 2m and 3m respectively. After the release of the blocks, the block "B" moves on the incline by 10 cm, before it is stopped. Neglecting friction between all surfaces, the displacement of wedge "C", in the meantime, is (in cm):

Figure 6
Blocks and wedge system
 Blocks and wedge system  (lmsq8.gif)

(a) 5 (b) 5 2 (c) 5 2 2 (d) 10 (a) 5 (b) 5 2 (c) 5 2 2 (d) 10

Solution

This question makes use of the concept of COM to a great effect. Answer to this question involving force analysis would have been difficult. Here, we note that The system of blocks and wedge is at rest in the beginning. This means that COM of the system is at rest before the block is released. When the blocks are released, the system is subjected to external force due to gravity. This means that COM should move along the direction of applied external force. As the direction of external force is vertically downward, the COM will also be accelerated downward. Importantly, COM will have no displacement in horizontal direction. We shall make use of this fact in solving this problem.

We can estimate here that the heavier block "B" slides down, whereas lighter block "A" slides up the incline. It means that block "B" moves towards right. In order that the COM has no horizontal displacement, we can conclude that wedge "C" moves towards left. Let us consider that the wedge "C" moves to left by a distance "x".

Figure 7
Free body diagram
 Free body diagram  (lmsq9.gif)

Now, there is yet another twist in the question. The displacement of block "B" is given with respect to wedge "C". The horizontal component of this displacement with respect to wedge ,d, is :

d = cos 45 0 = 5 2 d = cos 45 0 = 5 2

The horizontal component of displacement of the block "B" with respect to ground, d', is :

d' = d - x = 5 2 - x d' = d - x = 5 2 - x

As COM has no displacement in horizontal direction,

x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 = 0 x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 = 0

m 1 x 1 + m 2 x 2 = 0 m ( 5 2 - x ) - 3 m x = 0 m 1 x 1 + m 2 x 2 = 0 m ( 5 2 - x ) - 3 m x = 0

x = 5 2 2 x = 5 2 2

Hence, option (c) is correct.

Exercise 8

Two blocks of equal mass are released over a fixed double incline as shown in the figure. Neglecting friction between incline and blocks, the acceleration of the COM of the blocks and double incline system is :

Figure 8
Blocks and double incline system
 Blocks and double incline system  (lmsq5.gif)

(a) 2 - 1 2 3 g (b) 2 + 1 4 3 g (c) 3 - 1 4 3 g (d) 3 + 1 2 3 g (a) 2 - 1 2 3 g (b) 2 + 1 4 3 g (c) 3 - 1 4 3 g (d) 3 + 1 2 3 g

Solution

Each of the block behaves as particle. Thus, the acceleration of the COM of two blocks connected with "mass-less" string is given by :

a COM = m 1 a 1 + m 2 a 2 m 1 + m 2 a COM = m 1 a 1 + m 2 a 2 m 1 + m 2

The two blocks have same mass. Thus,

a COM = ( a 1 + a 2 ) 2 a COM = ( a 1 + a 2 ) 2

The two blocks are connected by a taut string having acceleration of same magnitude, say, "a". Now, as individual accelerations are at right angles, the magnitude of resultant acceleration is :

| a 1 + a 2 | = ( a 2 + a 2 = a 2 | a 1 + a 2 | = ( a 2 + a 2 = a 2

Thus, magnitude of acceleration of the COM is :

a COM = a 2 2 = a 2 a COM = a 2 2 = a 2

We need to find the common acceleration of the individual blocks by force analysis :

Figure 9
Free body diagram
 Free body diagram  (lmsq6.gif)

For first block :

F x = m g sin 60 0 - T = m a F x = m g sin 60 0 - T = m a

For second block :

F x = T - m g sin 30 0 = m a F x = T - m g sin 30 0 = m a

Solving two equations, we have :

a COM = g ( sin 60 0 - sin 30 0 ) 2 a COM = g ( sin 60 0 - sin 30 0 ) 2

a = 3 - 1 4 a = 3 - 1 4

Substituting this value in the equation of acceleration of COM, we have :

a = 3 - 1 4 3 g a = 3 - 1 4 3 g

Hence, option (c) is correct.

Answers


1. (d)    2. (b)     3. (d)     4. (a)     5. (a) 
6. (a)    7. (c)     8. (c)

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