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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