Question 449

Solution

Initial speed of the three-wheeler, *u* = 36 km/h = 10 m/s

Final speed of the three-wheeler, Time,

Mass of the three-wheeler,

Mass of the driver,

Total mass of the system,

According to the first law of motion,

Acceleration (

Therefore,

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,

= 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

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, Acceleration due to gravity,

Using Newton’s second law of motion,

Net force (thrust) acting on the rocket is given by the relation,

= 20000 × (10 + 5)

= 20000 × 15

= 3 × 10

Question 451

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 *t *= 0, the position of the body at that time to be *x *= 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, Mass of the body,

Force acting on the body,

Acceleration produced in the body, a =

(i)

At

Acceleration,

Initial velocity,

Using the relation,

s =

= 10 × (–5) = –50 m

(ii)

At

Acceleration,

USing the relation,

s' = ut' + (1/2) a" t

= 10 × 25 + (1/2) × (-20) × (25)

= 250 - 6250

= -6000 m

(iii)

At

For 0 ≤ t ≤ 30 s

Acceleration, a = -20 ms

Initial velocity, u = 10 m/s

Now, using the equation of motion, we have

s

= 10 × 30 + (1/2) × (-20) × (30)

= 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

= 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:

s

= -590 × 70

= -41300 m

∴ Total distance, s" = s

= -8700 -41300

= -50000 m

= -50 km.

Question 452

Solution

Given,

Initial velocity of the truck, *u* = 0

Acceleration, a = 2 m/s^{2}

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 (*v*_{x}) of velocity, in the absence of air resistance, remains unchanged.

i.e., * v*_{x} = 20 m/s

The vertical component (*v*_{y}) of velocity of the stone is given by the first equation of motion as,* v*_{y}* = u + a*_{y }δ*t *

where,

δ

a

∴

= 10 m/s

The resultant velocity (

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

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/s