Numerical Problems(Solved) : Force and Newton's Law of Motion, Science, Class 9 | EduRev Notes

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Class 9 : Numerical Problems(Solved) : Force and Newton's Law of Motion, Science, Class 9 | EduRev Notes

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. 
Sol. Here u = 0 m s-1; v = 30 ms-1; t = 10 s; a = ?
 Using v = u +at, we have 
 30 = 0+ a (10) 
 a = 3 m s-2 
 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 ?  
Sol. Here m = 70 g = 0.070 kg; u = 0.5 m s-1; v = 0; t = 0.5 s 
 F = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes 
 or F = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes 
 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 = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes = 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?  
Sol. Here m = 10 g = 0.010 kg ; u = a ; v = 300 m s-1 
 t = 0.003 s, F = ? 
Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes 
 or Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes 
 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 ?  
Sol. Here m = 1 kg; a = 1 ms-2 ; F = ? 
 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 
Sol. Here v _ u = 100 m s-1, m = 20 kg; F = 100 N ;t = ? 
 We know F = ma = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes 
 or t = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes = 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. 
 
Sol. (i) Here u = 20 m s_1; v = 0; s = 50 m; a = ? 
Using v2 - u2 = 2as, we have 
a = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes = _4 ms-2 
(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 
Sol. In first case m1 = 10 g = kg = 0.010 kg; 
 Now a1 = 5 ms-2 ; F1 = ? 
 F1 = m1a1 = 0.010 × 5
 F1 = 0.050 newton 
 In second case, m2 = 20 g =0.020 kg 
 or m2 = 0.020 kg 
 a2 = 2 m s-2 ; F2 = ? 
 Now F2 = m2a2 = 0.020 × 2 
 or F2 = 0.04 newton 
 We find that F1 > F2, hence more force is required to accelerate 10 g at 5 m s-2 than accelerating 20 g at 2 ms-2.  

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 at2, 
 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 m1, an acceleration of 8 ms-2 and a mass m2, an acceleration of 24 m s-2. What acceleration would it give if both the masses are tied together?  
Sol. Let us first find mass m1 and m2. 
 F = m1 a1 
 5 =m1 (8) or m1 = 5/8 kg 
 F = m2 a2 
 5 = m2 (24) or m2 = 5/24 kg 
 Total mass M = m+ m2 
 or M = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes
 

or M = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes kg

Let A be the acceleration produced in mass M. 
 F = MA 
 or 5 = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev NotesA 
 or A = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes = 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 ?  
Sol. Here m = 1,000 kg 
 u = 40 × ms-1 = ms-1 
 v = 0 ; t = 5 

 or F = 200 Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes

or F = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes

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/(105) 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 ms1. Calculate the velocity of recoil of the gun.
Sol. Mass of bullet, m = 100 g = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Noteskg = kg 
 Velocity of bullet, u = 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. 
 = 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.  
Sol. Here, Initial velocity of sphere, u = 0 
 Distance travelled, 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 
 v2 - u2 = 2as 
 v2 - 0 = 2 × 10 ms-2 × 0.8 m = 16 m2s-2 
 or u = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes = 4 ms-1.

Momentum of the sphere just before it touches the ground = mu = 10 kg × 4 ms1 = 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.  
Sol. Here, m1 = 10 g = kg = 102 kg 
 m2 = 20 g = 2 × 102 kg 
 u1 = 3 ms-1 ; u2 = 2 ms-1 
 u1 = 2.5 ms-1 ; u2 = ? 
 Momentum of first sphere before collision = m1u1 = 102 kg × 3 ms-1 = 3 × 102 kg ms-1 
 Momentum of second sphere before collision = m2u2 = 2 × 102 kg × 2 ms1 = 4 × 102 kg ms1 
 Total momentum of both the spheres before collision = m1u1  +m2u2 
 = 3 × 102 kg m/s + 4 × 102 kg m/s = 7 × 102 kg m/s
 

Now, momentum of first sphere after collision = m1u1 
 = 102 kg × 2.5 m/s = 2.5 × 102 kg m/s

Momentum of second sphere after collision = m2u2 
 = 2 × 102 kg × u2 m/s = 2 × 102 u2 kg m/s
 Total momentum of both the spheres after collision=m1u1 + m2u 2=2.5× 102 kgms-1 +2× 102u2kgms-1 
 Now, according to the law of conservation of momentum. 
 Total momentum after collision = Total momentum before collision
 2.5 × 102 + 2 × 102 u2 = 7 × 102
 or 2 × 102 u2 = 7 × 102 2.5 × 102
 = 4.5 × 102
 u2 = Numerical Problems(Solved) : Force and Newton`s Law of Motion, Science, Class 9 | EduRev Notes
 

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 ms1. They collide with each other and stick to each other after collision. What is the common velocity of these bodies after collision ?  
Sol. Let one body is moving towards left side and and second body is moving to the right side. So velocity of the body to the left side is taken as negative and velocity of body moving to the right side is taken as positive. 
 Momentum of body moving to left side before collision = mu = 0.5 kg × (2 ms-1) = 1 kg ms-1.

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