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A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

The given question can be illustrated as in the following fig. below.



Here, 

AO = incident path of the ball, 

OB = Path followed by ball after deflection, 

< AOB = Angle between the incident and deflected paths of the ball = 45o

Therefore, 

∠AOP = ∠BOP = 22.5° = θ

Initial velocity of the ball = Final velocity of the ball = v

On resolving the component of velocity along v, we have

Horizontal component of the initial velocity = vcos θ, along RO 

Vertical component of the initial velocity = vsin θ, along PO

Horizontal component of the final velocity = vcos θ, along OS

Vertical component of the final velocity = sin θ,  along OP

The horizontal components of velocities suffer no change.

The vertical components of velocities are in the opposite directions.

So, change in linear momentum of the ball gives us the impulse which is imparted to the ball. 

That is, 

Impulse = mvCosθ - (-mvCosθ) 

             =  2
mvCosθ 

Mass of the ball, m = 0.15 kg 

Velocity of the ball, v = 54 km/h

                                = 15 m/s


Therefore, 

Impulse = 2 x 0.15 x 15 cos 22.5o

             
= 4.16 kg m/s 

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A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s–1, what is the recoil speed of the gun?

Given, 

Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kg

Muzzle speed of the shell, v = 80 m/s

Recoil speed of the gun = V

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0

Final momentum of the system = mv – MV

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.

Now, as per the law of conservation of momentum. we have

Final momentum = Initial momentum 
 
          mv – MV = 0 

∴                   V = mv / 

                       = 0.020 × 80 / (100 × 1000)

                       = 0.016 m/s
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A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?


We have, 

Mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

Number of revolution per second, n =

40 / 60 = 2 / 3 rps

Angular velocity, ω =  = 2πn 

The tension in the string provides the centripetal force. 

i.e.,

T = Fcentripetal 

   =  

    = 
mrω

    = mr(
2πn)

    = 0.25 × 1.5 × (2 × 3.14 × ()2
    = 6.57 N 

Maximum tension in the string,


         
Tmax = 200 N 

           max = mv2max / r

∴       vmax = (Tmax × r  / m)1/2 

                 = (200 × 1.5 / 0.25)1/2 

                 = (1200)1/2 

                 = 34.64 m/s 

Therefore, the maximum speed of the stone is 34.64 m/s. 
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Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?


Between x = 0 and x = 2 cm, ball will be rebounding between two walls.

After every 2 s,

Magnitude of the impulse received by the ball = 0.08 × 10
–2 kg m/s

The given graph shows that a body changes its direction of motion after every 2 s.

Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions 
x = 0 and x = 2 cm.

The slope of the position-time
 graph reverses after every 2 s, the ball collides with a wall after every 2 s.

Therefore, ball receives an impulse after every 2 s.

Given, 

Mass of the ball, m = 0.04 kg

The slope of the graph gives the velocity of the ball.

Using the graph,

Initial velocity, (
u) = 
Velocity of the ball before collision, u = 10–2 m/s

Velocity of the ball after collision, v = –10–2 m/s 

Here, the negative sign arises as the ball reverses its direction of motion.

Magnitude of impulse = Change in momentum 

                                 = | mv - mu |

                                 = | 0.04 (v - u) |

                                 = | 0.04 (-10-2 - 10-2) |

                                 = 0.08 × 10-2 kg m/s 
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If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: 

(a) the stone moves radially outwards, 

(b) the stone flies off tangentially from the instant the string breaks,

(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?


(b) The stone flies off tangentially from the instant the string breaks.

When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks. 
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