Laws of Motion

Question 449
CBSEENPH11020066

The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg. 

Solution
Initial speed of the three-wheeler, u = 36 km/h = 10 m/s
Final speed of the three-wheeler, v = 0 m/s
Time, t = 4 s
Mass of the three-wheeler, m = 400 kg
Mass of the driver, m' = 65 kg
Total mass of the system, M = 400 + 65 = 465 kg
According to the first law of motion,
Acceleration (a) of the three-wheeler is, 
                       v = u + at 

Therefore,
 a = fraction numerator left parenthesis straight v space minus space straight u right parenthesis over denominator straight t end fraction space equals space fraction numerator left parenthesis 0 minus 10 right parenthesis over denominator 4 end fraction space equals space minus space 2.5 space straight m divided by straight s squared
The negative sign indicates that the velocity of the three-wheeler is decreasing with time.
Now, using Newton’s second law of motion, the net force acting on the three-wheeler is, 
       F = Ma

         = 465 × (–2.5)
 
         = –1162.5 N 
The negative sign indicates that the force is acting against the direction of motion of the three wheeler. 
Question 450
CBSEENPH11020067

A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s–2. Calculate the initial thrust (force) of the blast.
 

Solution
Mass of the rocket, m = 20,000 kg
Initial acceleration, a = 5 m/s2

Acceleration due to gravity, g = 10 m/s2

Using Newton’s second law of motion,
Net force (thrust) acting on the rocket is given by the relation, 
        F – mg = ma 

                 F = m (g + a)
                    = 20000 × (10 + 5)
                    = 20000 × 15
                    = 3 × 105 N
Question 451
CBSEENPH11020068

A body of mass 0.40 kg moving initially with a constant speed of 10 m s–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be = 0, the position of the body at that time to be = 0, and predict its position at t = –5 s, 25 s, 100 s.
 

Solution
Given,
Mass of the body, m = 0.40 kg
Initial speed of the body, u = 10 m/s, due north
Force acting on the body, F = –8.0 N
Acceleration produced in the body, a = straight F over straight m space equals space fraction numerator negative 8 over denominator 0.40 end fraction space equals space minus 20 space m divided by s squared
(i)
At t = –5 s,
Acceleration, a' = 0
Initial velocity, u = 10 m/s 
Using the relation, 
         s = ut + (1/2) a' t

           = 10 × (–5) = –50 m
(ii)
At t = 25 s
Acceleration, a'' = –20 m/s2 
Initial velocity, u = 10 m/s
USing the relation, 
            s' = ut' + (1/2) a" t2

               = 10 × 25 + (1/2) × (-20) × (25)

               = 250 - 6250
               = -6000 m
(iii)
At t = 100 s, 
For 0 ≤ t ≤ 30 s 
Acceleration, a = -20 ms-2
Initial velocity, u = 10 m/s
Now, using the equation of motion, we have
               s1 = ut + (1/2)a"t

                   = 10 × 30 + (1/2) × (-20) × (30)2

                   = 300 - 9000 
                   =  -8700 m
For 30 < t ≤ 100 s,
For t= 30 sec, as per the first equation of motion final velocity is given as, 

                   v
 = u + at 

                     = 10 + (–20) × 30
                     = –590 m/s 
Velocity of the body after 30 s = –590 m/s 
For motion between 30 s to 100 s, i.e., in 70 s: 
s2 = vt + open parentheses 1 half close parentheses a" t

    = -590 × 70
    = -41300 m 
∴ Total distance, s" = s1 + s2 
                          = -8700 -41300
                          = -50000 m
                          = -50 km.
Question 452
CBSEENPH11020069

A truck starts from rest and accelerates uniformly at 2.0 m s–2. At = 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 = 11 s? (Neglect air resistance.)

Solution

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 + aδ

where,
δt = 11 – 10 = 1 s, and 
a
y = g = 10 m/s

∴           vy = 0 + 10 × 1
                  = 10 m/s
The resultant velocity (v) of the stone is given as:
straight v space equals space square root of straight v subscript straight x squared space plus space straight v subscript straight y squared end root space

space space space equals space space square root of 20 squared plus space 10 squared end root space

space space space equals space 22.36 space straight m divided by straight s

Let, straight theta be the angle made by the resultant velocity with the horizontal component of velocity vx,
Therefore, 
tan space straight theta space equals space begin inline style straight v subscript straight y over straight v subscript straight x end style

straight theta space equals space tan to the power of negative 1 end exponent space open parentheses 10 over 20 close parentheses

space space space equals space 26.57 to the power of straight o 
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. 

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