Question

*Q.02.*
A car of mass 1600kg moving with the initial velocity of 18 m/s, hits another stationary car of mass 1400kg, and they lock together, with what velocity they move after collision??

*Q.03.*
A car weighing 9800N is moving at 25 m/s, if the frictional force is acting on it is 2000N, how fast is the car moving when it has travelled 60 m.??

*Q.04.*
Two bodies of mass *M* and *m* are connected to the ends of a string which passes from a frictionless pulley such that two bodies move vertically, derive expression for,
a) Acceleration of two bodies
b) Tension of two bodies
c) if M is 3kg what will be the acceleration?

*Q.05*
What do you about the laws of motion, explain in details with proper examples and formulas.

*Q.06*
What are the equations of motion?? Write formulas also derive at least one equation of motion.


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Answer 1

Answer:

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Answer 2

Answer:

1. The velocity after the collision is 9.6m/s.

Given:

Moving a car of mass

M1 = 1600kg

Mass of stationery car

M2 = 1400kg

Initial velocity u is 18 m/s.

To Find :

V =?

Solution:

V1 = V2 =V

We know that to find the velocity we can use the following equation as,

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

1600×18 + 1400×0 = 1600v + 1400v

28800 = 3000v

$v = \frac{28800}{3000}$

v = 9.6 m/s.

Hence the velocity after the collision is 9.6m/s.

2. The car moved at 19.6 m/s when it travelled 60m.

Given:

W = 9800N

F = 2000N

S= 60 m

u1 = 25m/s

To Find :

How fast the car moves when it travels 60m?

Solution:

${ u_{f}}^{2} = { u_{i} }^{2} + 2as$

We know that,

F = ma

Mass is given by,

$\frac{9800}{9.8}$

= 1000kg

Rearrange equation one as

$a = \frac{ u_{f}^{2} - 625}{120}$

and the second equation as

F = mamoved= 1000 × a

$1000 { u_{f}}^{2} = 625000 - 240000$

${ u_{f} }^{2} = \sqrt{385}$

= 19.6 m/s

Hence, the car moved at 19.6 m/s when it travelled 60m.

3.

Given:

Two bodies having mass M and m.

To Find :

• a)Acceleration of two bodies
• b) Tension of two bodies
• c) if M is 3kg what will be the acceleration

Solution:

We know that Forces are given by

T - Mg = Ma For mass M ----------(1)

T - mg = mg For mass m ----------(2)

T = mg + ma

put this equation in equation 1 we get

mg + ma - Mg = Ma

g(M - m) = a (M+ m)

$a = (\frac{ M - m }{ M + m} )g$

$T = m (\frac{ M -m }{ M + m } )g + mg$

$T= \: \frac{ \: M m-{m}^{2} }{ M + m} g + mg</p><p>$

$T = g(\frac{ 2Mm}{ M + m } )$

a =( 3 - m/3+m )×g

$a \: = 3g( \frac{ Mm}{ M + m } )$

Hence Proved.

4. Three laws of motion are as follows

a) Newton's first law is the law of inertia.

• It states that when the body is at rest or moving at a constant speed it will remain at rest or moving at a constant speed until a force act on it.

b) Newton's second law states that the force acting on the body is equal to the products of its mass and acceleration

F = ma

c) Newton's third law is about every action that has equal and opposite reactions.

5. The equations of motion are as follows

• v = u + at
• v² = u² + 2as
• s = ut + ½at²

equation 3 is derived as follows

Let S be a distance

We know that

Distance = Average velocity × Time.

Also, the Average velocity (u+v)/2

Distance (s) = (u+v)/2 × t

Also, from v = u + at

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

or

s = ut +½ at²

Hence proved.

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