A truck starts from rest and accelerates uniformly at 2.0 m s–
zigya tab
Advertisement

A truck starts from rest and accelerates uniformly at 2.0 m s–2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)


Given, 

Initial velocity of the truck, u = 0

Acceleration, a = 2 m/s2

Time, t = 10 s

a)

According to the equation of motion, we have

                   v = u + at

                     = 0 + 2x10

                     = 20 m/s

The final velocity of the truck and hence, of the stone is 20 m/s.

At t = 11 s, the horizontal component (vx) of velocity, in the absence of air resistance, remains unchanged.

i.e.,                vx = 20 m/s 

The vertical component (vy) of velocity of the stone is given by the first equation of motion as,

                     vy = u + ay δt 

where,

δt = 11 – 10 = 1 s, and 

a
y = g = 10 m/s2 

∴           vy = 0 + 10 × 1

                  = 10 m/s

The resultant velocity (v) of the stone is given as:




Let,  be the angle made by the resultant velocity with the horizontal component of velocity vx,

Therefore, 

 

b) 

When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s2 and it acts vertically downward. 

290 Views

Advertisement
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s–1. What is the trajectory of the bob if the string is cut when the bob is
(a) at one of its extreme positions,
(b) at its mean position.

(a) When bob is at its extreme position:

The trajectory of the bob if the string is cut at one of it's extreme positions will be in v
ertically downward direction.

At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground. 

(b) When the bob is at its mean position:

When the string is cut at its mean position, the trajectory of the bob will be a parabolic path. 
At the mean position, the velocity of the bob is 1 m/s.

The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only.

Hence, the bob will follow a parabolic path. 
361 Views

A man of mass 70 kg stands on a weighing scale in a lift which is moving, 
(a) upwards with a uniform speed of 10 m s–1, 

(b) downwards with a uniform acceleration of 5 m s–2, 

(c) upwards with a uniform acceleration of 5 m s–2.
 
What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?


(a) 

Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton’s second law of motion, we can write the equation of motion as, 

                      R – mg = ma

where,

 
ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

∴          R = mg 

              = 70 × 10

              = 700 N 

Therefore, reading on the weighing scale =

                                                     =

                                                     =


(b) 

Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 , downward

Using Newton’s second law of motion, we can write the equation of motion as: 

                      R + mg = ma 

                               R = m(g – a) 

                                  = 70 (10 – 5)

                                  = 70 × 5 

                                          = 350 N 

∴ Reading on the weighing scale = 350 g = = 35 kg

(c) 

Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 upward 

Using Newton’s second law of motion, we can write the equation of motion as:

                        R
 – mg = ma 

                                 R = m(g + a) 

                                    = 70 (10 + 5)

                                    = 70 × 15

                                    = 1050 N
Therefore,

Reading on the weighing scale =

                                               =

                                               = 105 kg


(d) When the lift moves freely under gravity, acceleration a =g 

Using Newton’s second law of motion, we can write the equation of motion as, 

                       R + mg = ma 

                                 R = m(g – a) 

                                    = m(g – g)

                                    = 0 

∴ Reading on the weighing scale = = 0 kg 

The man will be in a state of weightlessness.
916 Views

Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t> 4 s,0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only). 


(a)

For t < 0

From the given graph, the position of the particle is coincident with the time axis.

That is, the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.


For t > 4 s 

In the given graph, the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of  
3 m from the origin. Hence, no force is acting on the particle.

For 0 < t < 4 

The position-time graph has a constant slope in the given graph.

Hence, the acceleration produced in the particle is zero.

Therefore, the force acting on the particle is zero.

(b)

At t = 0,

Impulse = Change in momentum 

          = mv – mu 

Mass of the particle, m = 4 kg 

Initial velocity of the particle, u = 0 

Final velocity of the particle, v =  m/s

∴ Impulse = 4 x (  - 0)

              = 3 kg m/s


At t = 4 s,

Initial velocity of the particle, u =  m/s 

Final velocity of the particle, v = 0 

∴ Impulse = 4 (0 -  ) = -3 kg m/s 
(a) For t < 0

It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.

For t > 4 s 

It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 
3 m from the origin. Hence, no force is acting on the particle.

For 0 < t < 4

It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.

(b) At t = 0

Impulse = Change in momentum 

           = mv – mu 

Mass of the particle, m = 4 kg 

Initial velocity of the particle, u = 0 

Final velocity of the particle, v =  m/s 

∴ Impulse = 4 (  - 0) = 3 kg m/s

At t = 4 s, 

Initial velocity of the particle, u =  m/s

Final velocity o9f the particle, v = 0

∴ Impulse = 4 (0 - ) = -3 kg m/s 
150 Views

Two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 m/s collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Mass of billiard balls = 0.05 kg 

After collision the two balls rebound with same speed, therefore, the change in the momentum of each ball is, 

                     increment p equals 2 m u 

Substituting m = 0.05 kg and u = 6 m/s.

We get,

Impulse, I italic equals italic increment p space equals space 2 cross times 0.05 cross times 6 equals 0.6 space Ns




 
232 Views

Advertisement