So the direction of the force will be west to east. retardation) hence the direction of the force would be opposite to the initial direction. As the velocity reduces because of force, (negative acc i.e. acceleration = change of vel/time=5/2 = 2.5 m/s 2įorce = mass X acceleration= 1 X 2.5 = 2.5 N. Mass = 1kg velocity = 10 m/s (east to west) Because of the force the velocity reduces to 5 m/s in 2 secs. What is the value and direction of the force? A force is applied to it for 2 seconds and its velocity becomes 5 m/s. Mass = force/acceleration here acceleration =a = (10-5)/1 = 5 m/s 2 so, mass = 200/5 kg = 40 kgħ) A mass of 1 kg is moving from east to west with a velocity of 10 m/s. Note:Force = ma = m(v-u)/t = (mv – mu)/t = change of momentum /time = rate of change of momentumįorce= change of momentum /time = (200-100)/2 = 100/2 = 50 NĦ) A force of 200 N is applied to a body and its velocity changes from 5m/s to 10m/s in a second. Physics problems for class 9 with solutions – physics problems solvedĪcceleration = change in velocity / time = 32/0.01 = 3200 m/s 2įorce = mass X acc = 0.15 X 3200 N= 480 N.ĥ) Momentum of an object changes from 100 kg m/s to 200 kg m/s in 2 seconds. What was the force applied to the ball when it hits the wall? (350/800) 2 = 175 – 76.5 = 98.5 mĤ) A ball of 150 g mass moves with 12 m/s and bounces back after hitting a wall with 20 m/s after a small duration of 0.01 second. Mass = 50 kg u = 400 m/s v= 50 m/s F = 40000 N t = unknown Distance travelled in time t =?Īcceleration = a = F/m = 40000/50 = 800 m/s 2Īgain, acceleration a = change of velocity /t or, t = change of velocity/a = 350/800 sec What is the distance traveled by the mass during this period? A force of 40000 N is applied to the mass and its velocity is reduced to 50 m/s after some time. So the force applied on the mass for first 3 secs= mass x acc = 16 x 9 = 144 Nģ) A mass of 50 kg was moving with a velocity of 400 m/s. Therefore the acceleration in first 3 seconds caused by the force present = (27-0)/3 = 9 m/s². So in the first 3 seconds, velocity changes from 0 to 27m/s. This velocity was attained in the first 3 seconds when the force was present. So the velocity in the next 3 secs: V = 81/3 = 27 m/s In the last 3 seconds, the mass travels with uniform velocity (as there is no force that means no acceleration). What was the value of the force applied?įirst 3 secs: Force is present, and in the next 3 secs Force is not there, so uniform velocity for the last 3 seconds mentioned. As the force is removed the mass moves 81 meters in 3 seconds. Solution: F =1000 N m=25kg t= 5 sec u=0 v=?Īcceleration = a = F/m = 1000/25 = 40 m/s 2Ģ) A force is applied to a mass of 16 kg for 3 seconds. Solved Numerical Problems in Physics class 9 motion & forceġ) A force of 1000 Newton is applied on a 25 kg mass for 5 seconds. Suggested for CBSE, ICSE, state boards, IGCSE, GCSE, UPSC, and SSC examinations. Most of these questions & numerical are with on-page solutions or links to the solution pages. The following sections contain more than a hundred Solved Numerical Problems in Physics for class 9 covering multiple important chapters including motion physics (Kinematics).
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