**Subtraction of vectors
**

Since, and

Since,

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But and

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**Step 1 : **Read the question carefully and visualise the situation (think as if the process is happening in front of you).

**Step 2 :** Draw a relevant figure and write the parameters given.

**Step 3 :** Make sure all values are in the same set of units.

**Step 4 : **Plan the concept/ formulae which are to be used.

**Step 5 :** Avoid doing calculations in each step. Try to do all the calculations at final step to save time.

*Solution: *

Here, m = 2000 kg; a = – 1.5 ms^{–2}; F = ?

Using, F = *ma*, we get

F = 2000 kg (– 1.5) ms ^{–2 }= – 3000 kg ms^{– 2}

**= – 3000 N **

Negative sign shows that force of friction (F) acts in a direction opposite to the motion of the motorcycle.

*Solution: *

Given u = 3 ms^{–1}, *v* = 7 ms^{–1}, t = 2s, m = 5 kg

Also F = *ma *

*u* = 3 ms^{–1}, *t* = 5 s, a = 2 ms^{– 2}, *v* = ?

Applying, *v *= *u* + *at*

**Solution: **

Here, mass *m* = 5000 kg; *u* = 0; s = 200 m; t = 10 s;

*a* = ?; F = ?

or *a* = 4 ms^{ – 2}.

Using , F = *ma*, we get

F = 5000 kg 4 ms^{– 2 }= 20,000 kg ms^{– 2}

**= 20,000 N **

*Solution***: **

The velocity-time graph shows that the velocity of the ball at *t* = 0 is zero. That is initial velocity of ball, *u* = 0

Velocity of ball at t = 4 s is 20 ms^{– 1}

That is, final velocity, *v* = 20 ms ^{– 1}

time, *t = *4 s

*a* = ? ; F = ?

**Step 1.** Acceleration of the ball *a* =

**Or a = 5 ms^{– 2} **

**Step 2. **Also, mass of ball, *m *= 20 g = kg =kg

Force acting on the ball, F = *ma*

= 0.1 N

*Solution: *

The velocity-time graph shows that velocity of the ball at *t* = 0 is 30 ms^{–1}

That is, initial velocity of the ball, *u* = 30 ms^{–1}

The velocity of the ball at *t* = 6s is zero.

That is, final velocity of the bal, = 0

time, *t* = 6s ; F = ?

**Step 1.** Acceleration of the ball, *a *=

= – 5 ms^{–2}

Negative sign shows that the ball is retarded or decelerated.

**Step 2.** Also, mass of ball, m = 50 g

Force acting on the ball, F = *ma*

*= ***– 0.25 N **[ 1 kg ms^{ – 2} = 1 N]

*Solution. *

Here, m = 1 kg.; u = 20 ms ^{– 1}; ; s = 50 m; F = ?

= – 4 ms^{– 2}

**Step 2.** Using F = *ma*, we get

= – 4 kg ms^{– 2} = **– 4N **

Negative sign shows that the force is retarding force.

*Solution. *

**For First Block: **m = m_{1} kg.; *a *= 10 m/s^{2}

Using F = *ma*, we get

**For Second Block: **

m = *m*_{2} kg; *a* = 20 ms^{– 2}; F = 5 N

Using, F = *ma*, we get

**When both the blocks are tied together **

Mass of the combination, *m* = (*m*_{1} + *m*_{2}) = ( 0.5 kg + 0.25 kg) = 0.75 kg

F = 5 N

Using F = *ma*, we get

**Solution. **

Here, *m* = 1000 kg

Initially velocity, *u* = 36 km/h =

Time, *t* = 5 s.

F = ?

Using, F = *ma*, we get

Thus force exerted by the car on the wall = 2000N

**Solution. **

Here, mass of ball, m = 100g =kg = 0.1 kg

Initial velocity of ball, *u* = 10 ms^{–1}

Time, *t* = 0.2 s.

F = ?

step 2. Using, **F = ma, we get **

F = 0.1 kg ( – 50 ms^{– 2}) = – 5 kg ms^{– 2 }**= – 5 N **

Negative sign shows that force applied by the boy is the opposing force *i*.*e*. a force in a direction opposite to the direction of motion of the ball.

**Solution. **

Here, initial velocity, u = 36 km h^{–1} = ms^{–1}= 10 ms^{–1}

Time *t* = 5 s.

Mass of car, *m* = 1000 kg

Force required to stop the car,

F = ma = 1000 kg (– 2 ms^{ – 2}) = **– 2000 N**

**Setp 2.** Mass of the bus, *m* = 8000 kg

Force required to stop the bus,

F = ma = 8000 kg (– 2 ms^{– 2}) = –** 16000 N **

Negative sign with force shows that the force is applied in a direction opposite to the direction of the motion of the car and bus. Such a force is known as **retarding force**.

**Solution. **

F = 0.2 N

*t* = 5 s

*u = *0

S = ?

**Step 1.** F = *ma*, we get

**Step 2.** To find distance traveled, using

S = (2 ms^{–2}) (25 s)^{2} =** 625 m**

**Solution. **

*u* = 100 ms^{ – 1}

*t = *0.05 sec

= 0 (finally bullet comes to rest)

F = ? ; S = ?

(*i*) Using, F = *ma*, we get

= – 100 kg ms^{ – 2} = –100 N

Thus, force exerted by the wooden plank on the bullet = 100 N.

s = 100 ms^{–1} 0.05 s + ( 0 – 100 ms^{–1}) 0.05 s

= 5 m – 2.5 m = **2.5 m. **

**Solution. **

Here, *m* = 500 g = kg; *u* = 20 ms^{– 1}

Using, F = *ma*

Therefore, force exerted by nail on the hammer = 500 N

**Solution. **

Velocity of bullet, = 100 ms^{ – 1}

Mass of gun, M = 20 kg

Let recoil velocity of gun = V

**Step 1.**Before firing, the system (gun + bullet) is at rest therefore, initial momentum of the system = 0

Final momentum of the system = momentum of bullet + momentum of gun

**Step 2. **Apply law of conservation of momentum

Final momentum = Initial momentum

*i*.*e*. 10 + 20 V = 0

20 V = – 10

Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet.

**Solution. **

Here, Initial velocity of sphere, u = 0

Distance travels, S = 80 cm = 0.8 m

Acceleration of sphere, *a *= 10 ms ^{– 2}

**Step 1. **Final velocity of sphere when it just reaches the ground can be calculated using

= 16 m^{2} s^{– 2 }

= 4 ms^{–1}

Momentum of the sphere just before it touches the ground =

= 10 kg4 ms^{– 1} = 40 kg ms^{–1}

**Step 2. **On reaching the ground, the iron sphere comes to rest, so its final momentum = 0

According to the law of conservation of momentum,

Momentum transferred to the ground = momentum of the sphere just before it comes to rest

= 40 kg ms^{– 1}

*Solution*.

Here, *m*_{1} = 10 g = kg = 10^{– 2} kg

*m*_{2 }=_{ }20 g = 2 10^{– 2}kg

*u*_{1 }= 3 ms^{–1}; *u*_{2} = 2 ms^{– 1}

Momentum of first sphere before collision = *m*_{1}*u*_{1}

Momentum of second sphere before collision = *m*_{2}*u*_{2}

Total momentum of both the sphere before collision = *m*_{1}*u*_{1} + *m*_{2}*u*_{2}

= 3 10^{– 2} kg ms^{–1 }+ 4 10^{– 2} kg ms^{–1}

Now, momentum of first sphere after collision =

Momentum of second sphere after collision^{ }=

Total momentum of both the spheres after collision

= 2.5 10^{– 2} kg ms^{– 1} + 210^{– 2} kg m^{–1}

Now, according to the law of conservation of momentum.

Total momentum after collision = Total momentum before collision

2.5 10^{– 2} + 2 10^{– 2} = 7 10^{– 2 }

Or 210^{– 2} = 710^{– 2} – 2.510^{– 2}

= **2.25 ms ^{–1} **

*Solution***. **

Let one body is moving towards left side and second body is moving to the right side. So velocity of the body to the left side is taken as positive and velocity of body moving to the right side is taken as negative.

Momentum of body moving to left side before collision = *mv *

= 0.5kg 2 ms^{–1} = 1 kg ms^{–1}

Momentum of body moving to right side before collision = – *mv*

= – 1 kg ms^{–1}

Total momentum of both the bodies before collision

= 1 kg ms^{–1 }– 1 kg ms^{–1 }= 0

Let V = common velocity of both the bodies after striking to each other

Total momentum of both the bodies after collision

= (*m*_{1} + *m*_{2}) V

= (0.5 + 0.5) V = V kg ms^{– 1}

According to the law of conservation of momentum

Total momentum after collision = Total momentum before collision

*i*.*e*., V = 0

Thus, the combination of two bodies comes to rest after collision.