An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Solution and Explanation
Mass of the object, m= 100 kg,
Initial velocity, u= 5 ms-1
Final velocity, v= 8 ms-1
(i) Initial momentum =mu= 100 x 5=500 kg-ms-1
Final momentum = mv = 100 x 8 = 800 kg-ms-1
(ii) From Newton’s second law, Force exerted on the object = Rate of change of momentum
= Change in momentum
= Time Final momentum- Initial momentum
Time = \(\frac{800-500}{6}\) =\(\frac{300}{6}\)= 50N
Top Questions on Inertia and Mass
- Two bodies of different masses are moving with the same kinetic energy. Which one has a greater momentum?
- CUET (PG) - 2023
- Physics
- Inertia and Mass
- What will be the expression for moment of inertia of a rectangular lamina (ABCD having length \(l\) and breadth b) about an axis passing through one of its corners and perpendicular to its plane?
- CUET (PG) - 2023
- Physics
- Inertia and Mass
- Match List I with List II
LIST-I LIST-II A. Green's theorem I. Moment of inertia B. Kepler's laws II. Vectors C. Theorem of parallel and
perpendicular axesIII. Inertia D. Newton's law IV. Motion of planets Choose the correct answer from the options given below:- CUET (PG) - 2023
- Physics
- Inertia and Mass
- An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
- CBSE Class IX
- Science
- Inertia and Mass
- A bullet of mass 10 g traveling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
- CBSE Class IX
- Science
- Inertia and Mass
Questions Asked in CBSE Class IX exam
- Explain what the reasons for the following are.
(i) Einstein leaving the school in Munich for good.
(ii) Einstein wanting to study in Switzerland rather than in Munich.
(iii) Einstein seeing in Mileva an ally.
(iv) What do these tell you about Einstein?- CBSE Class IX
- A truly beautiful mind
- (i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)- CBSE Class IX
- The snake and the mirror
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.
- CBSE Class IX
- Angle Subtended by an Arc of a Circle
- Name the two temples the author visited in Kathmandu.
- CBSE Class IX
- Kathmandu
Look up the dictionary entries for the words sympathy, familiarity, comfort, care, and surprise. Use the information given in the dictionary and complete the table.
Noun, Adjective, Adverb, Verb, Meaning:
sympathy
familiarity
comfort
care
surprise- CBSE Class IX
- If I were you
Concepts Used:
Laws of Motion
The laws of motion, which are the keystone of classical mechanics, are three statements that defined the relationships between the forces acting on a body and its motion. They were first disclosed by English physicist and mathematician Isaac Newton.
Newton’s First Law of Motion
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion
Newton's 2nd law of motion deals with the relation between force and acceleration. According to the second law of motion, the acceleration of an object as built by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Newton’s Third Law of Motion
Newton's 3rd law of motion states when a body applies a force on another body that there is an equal and opposite reaction for every action.