The document Numerical Problems(Solved) : Force and Newton's Law of Motion, Science, Class 9 | EduRev Notes is a part of the Class 9 Course Science Class 9.

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**SOLVED EXAMPLES ****Ex.1** **Calculate the force required to impact to a car, a velocity of 30 m s ^{-1} in 10 seconds. The mass of the car is 1,500 kg. **

Using v = u +at, we have

30 = 0+ a (10)

a = 3 m s

Now F = ma = 1,500 × 3

or F = 4,500 N.

**Ex.2** **A cricket ball of mass 70 g moving with a velocity of 0.5 m s ^{-1} is stopped by player in 0.5 s. What is the force applied by player to stop the ball ? **

F =

or F =

or F = - 0.07 newton

**Ex.3** **What will be acceleration of a body of mass 5 kg if a force of 200 N is applied to it ? ** **Sol.** Here m = 5 kg; F = 200 N

F = ma or a = F/m

a = = 40 m s^{-2}

a = 40 m s^{-2}

**Ex.4** **A bullet of mass 10 g is fired from a rifle. The bullet takes 0.003 s to move through its barrel and leaves with a velocity of 300 ms ^{-1}. What is the force exerted on the bullet by the rifle? **

t = 0.003 s, F = ?

or

or F = 1,000 N.

**Ex.5** **What force would be needed to produce an acceleration of 1 ms ^{-2} on a ball of mass 1 kg ? **

Now F = m a = 1 × 1

or F = 1 newton.

**Ex.6** **What is the acceleration produced by a force of 5 N exerted on an object of mass 10 kg ? ** **Sol.** Here F = 5 N; m = 10 kg; a = ?

Now F = ma o

a = 0.5 ms^{-2}

**Ex.7 How long should a force of 100 N act on a body of 20 kg so that it acquires a velocity of 100 ms ^{-1} ? **

We know F = ma =

or t = = 20 s.

**Ex.8** **A 1,000 kg vehicle moving with a speed of 20 m s ^{-1} is brought to rest in a distance of 50 m,(i) Find the acceleration;(ii) Calculate the unbalanced force acting on the vehicle;(iii) The actual force applied by the brakes may be slightly less than that calculated in (ii). Why? Give reason. **

Using v

a = = _4 ms

(ii) F = ma = 1,000 × (-4) = -4,000 N

(iii) Due to force of friction, the actual force applied by brakes may be slightly less than calculated one.

**Ex.9** **Which would require greater force: accelerating a 10 g mass at 5 m s ^{_2} or 20 g mass at 2 m s^{-}^{2}? **

Now a

F

F

In second case, m

or m

a

Now F

or F

We find that F

**Ex.10** **A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric ton. ** **Sol.** Here u = 0 m s^{-1} (Starting from rest); s = 400 m; t = 20 s; a = ?

Using s = ut+1/2 at^{2},

We have 400 = 0 × a × 20 × 20 = 200 a

or a = 2 ms^{-2}

Now mass m = 7 metric ton = 7,000 kg; F = ?

F = m a = 7,000 × 2

or F = 14,000 N.

**Ex.11** **A force of 5 N gives a mass m _{1}, an acceleration of 8 ms^{-2} and a mass m_{2}, an acceleration of 24 m s^{-2}. What acceleration would it give if both the masses are tied together? **

F = m

5 =m

F = m

5 = m

Total mass M = m

or M =

or M = kg

Let A be the acceleration produced in mass M.

F = MA

or 5 = A

or A = = 6 ms^{-2}

Hence the acceleration of the combination is 6 ms^{-2}. **Ex.12** **A car of mass 1,000 kg moving with a velocity of 40 km h ^{_1} collides with a tree and comes to stop in 5 s. What will be the force exerted by car on the tree ? **

u = 40 × ms

v = 0 ; t = 5

or F = 200

or F =

or F = -2,222 N

**Ex.13** **The velocity-time graph of a coin moving on a floor is as given in Fig. How much force does the floor exert on the coin? Given the mass of the coin is 10 g. ** **Sol.** Here, u = 24 cm s^{-1}

v = 0 ms^{-1}; t = 8 s ; a = ?

v = u at

F = m a;

Here m = 10 g

F = 10 (-3) = - 30 dyne = - 30/(10^{5}) N

or F = -3 × 10^{-}^{4} N.

Negative sign tells us that the force is acting in the direction opposite to the direction of motion.

**Ex.14** **A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity fo 100 ms ^{1}. Calculate the velocity of recoil of the gun.**

Velocity of bullet, u = 100 ms-

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.

= mu+ MV = 100 + 20 V = 10+20 V**Step 2.** Apply law of conservation of momentum

Final momentum = Initial momentum

i.e. 10+ 20 V = 0

20 V = 10

or V = = 0.5 ms^{-}^{1}.

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

**Ex.15** **An iron sphere of mass 10 kg is dropped from a height of 80 cm. If the downward acceleration of the ball is 10 ms ^{-}^{2}, calculate the momentum transferred to the ground by the ball. **

Distance travelled, s = 80 cm = 0.8 m

Acceleration of sphere, a = 10 ms

v

v

or u = = 4 ms

Momentum of the sphere just before it touches the ground = mu = 10 kg × 4 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}.

**Ex.16** **Two small glass spheres of masses 10 g and 20 g are moving in a straight line in the same direction with velocities of 3 ms ^{-1} and 2 ms^{-}^{1} respectively. They collide with each other and after collision glass sphere of mass 10 g moves with a velocity of 2.5 ms^{-}^{1}. Find the velocity of the second ball after collision. **

m

u

u

Momentum of first sphere before collision = m

Momentum of second sphere before collision = m

Total momentum of both the spheres before collision = m

= 3 × 10

Now, momentum of first sphere after collision = m_{1}u_{1}

= 10^{2} kg × 2.5 m/s = 2.5 × 10^{2} kg m/s

Momentum of second sphere after collision = m_{2}u_{2}

= 2 × 10^{2} kg × u_{2} m/s = 2 × 10^{2} u_{2} kg m/s

Total momentum of both the spheres after collision=m_{1}u_{1} + m_{2}u_{ 2}=2.5× 10^{2} kgms^{-}^{1} +2× 10^{2}u_{2}kgms^{-}^{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} u_{2} = 7 × 10^{2}

or 2 × 10^{2} u_{2} = 7 × 10^{2} 2.5 × 10^{2}

= 4.5 × 10^{2}

u_{2} =

**Ex.17** **Two bodies each of mass 0.5 kg are moving in a straight line but opposite in direction with the same velocity of 2 ms ^{1}. They collide with each other and stick to each other after collision. What is the common velocity of these bodies after collision ? **

Momentum of body moving to left side before collision = mu = 0.5 kg × (2 ms

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