Skip to main content

Motion Numericals

CBSE ADDA

Q. 1. How much force is needed to accelerate a trolley of mass 20g through 1 m/s2?
Q. 2. A force of 100N acts on a mass of 25 kg for 5 s .What velocity does it generate?
Q. 3. A bullet leaves a rifle with a velocity of 100m/s and the rifle of mass 2.5 kg recoils with a velocity of 1m/s. Find the mass of the bullet?
Q. 4. A certain force acting on a mass of 15kg for 3s, gives it a velocity of 2m/s. Find the magnitude of force?
Q. 5. A cricket ball of mass 0.15 kg is moving with a velocity of 1.2m/s . Find the impulse on the ball and average force applied by the player if he is able to stop the ball in 0.18s?
Q. 6. A motor car of mass 200kg is moving with a certain velocity . It is brought to rest by the application of brakes, within a distance o f 20m when the average resistance being offered to it is 500N.What was the velocity of the motor car?
Q. 7. A body of mass 0.5kg undergoes a change of velocity of 4cm/s in 4s What is the force acting on it?
Q. 8. A force of 8N acting on an 8kg mass for 4s provides its some velocity. Calculate the velocity?
Q. 9. What would be the weight of a body of mass 50kg on the surface of the moon, where g = 1.6m/s2 what would be its mass?
Q. 10. A boy jumps a distance of 2m on the surface of the earth. What distance will he jump on the surface of the moon where g is 1/6th of the value on the surface of the earth?
Q. 11. Two bodies of mass 1kg and 2 kg moving in the direction opposite to each other with a speed 5m/s collide. Calculate the total momentum of the system before collision?
Q. 12. A bullet of mass 15g leaves the barrel of a gun with a velocity of 120m/s.The gun recoils with a velocity of 1m/s. Find the mass of the gun?
Q. 13. A force of 3N acts on a mass of 0.5kg at rest for 10s. Find the final velocity and the momentum of the body after 10s?
Q. 14. A force of 80N acting on a certain mass for 3s gives it a velocity of 6m/s. find the mass of the body if the body was initially at rest?
Q. 15. A toy of mass 0.1kg acquires a spped of 5 m/s when pushed forward. Ahat is the impulse given to the toy?
Q. 16. A projectile weighing 200kg is fired from a gun with a velocity of 500m/s. If the velocity of recoil of the gun is 12.5m/s calculate the mass of the gun?
Q. 17. A car of mass 2500kg accelerates at the rate of 4m/s2 and then at the rate of 16 m/s2. Calculate t6he ratio of the forces exerted by the engine in the two cases?
Q. 18. A driver accelerates his car first at the rate of 1.8m/s2 and then at the rate of 1.2m/s2 calculate the ration of the forces exerted by the engines?
Q. 19. A body of mass 5 kg undergoes a change in speed from 30 to 40m/s . calculate its increase in momentum?
Q. 20. A body of mass 100g is moving with a velocity of 15m/s Find the momentum associated with the ball?
Q. 21. body of mass 5 kg is moving with a uniform velocity of 10m/s It is acted upon by a force of 20N What will be its velocity after 1s? Q. 22. In order to gain a velocity of 12m/s how long should a force of 30N be exerted on a body of mass 5kg? Q. 23. The velocity of a body of mass 5kg reduces from 10 m/s to 5 m/s. What is the change in momentum of the body?
Q. 24. A 5 kg rifle fires a 10g bullet at a velocity of 500m/s find the recoil velocity of the rifle?
Q. 25. A block of mass 3kg has a velocity of um/s. When a force of 18N acts on the block, it reduces the velocity from U to U/2 after the block has covered a distance of 9 m . Find U?

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get d...

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page ...

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution:  Mass of the reactants = 10g ;  Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g.  In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 ...