A force of 0.6 gf acts on a glass ball of mass 200 g for 12 s. If initially the ball is at rest, calculate the final velocity and distance covered. (Take g = 1000 cms–2)
Answers
SOLVED NUMERICAL PROBLEMS
1.A body having a velocity of 200 ms-1 has a momentum of 5 Ns. Find the mass of body.
Ans: m=? v=200ms-1
Momentum = 5 Ns.
Momentum = m x v.
5 Ns = m x 200 ms-1 \m =0.025 kg.
2 Calculate the momentum of an electron of mass 9 x 10-31 kg. Moving with a velocity of 6 x 107 ms-1.
Ans: Momentum = ? m= 9 x 10-31 kg; v=6 x 107 ms-1
Momentum = m x v = 9 x 10-31 kg x 6 x 107 ms-1=54 x 10-24Ns.
3. What force will produce an acceleration of 3.5 ms-2, in a body of mass 12.5 kg?
Ans: F=? a = 3.5 ms-2 m=12.5 kg.
F=ma = 12.5 kg x 3.5 ms-2 = 43.75 N.
Calculate the mass of a body. When a force of 525 N, produce an acceleration of 3.5 ms-2.
Ans: m=? F= 525 N; a=3.5 ms-2
M=F ¸ a =525 ¸3.5 = 150 kg.
A force of 50 kgf acts on a body of mass 1/2 tonne, find the acceleration produced in body. [Take g = 10 Nkg-1]
Ans: F = 50 kgf = 50 x 10 =500N, m=1 x 1000 = 500 kg; a=?
F 500
A = = =1 ms-2.
M 500
What is the mass of a body, which is moving with an acceleration of 1.4 ms-2, when a force of 50 kgf acts on it? [g=9.8 ms-2]
Ans: F=50kgf = 50 x 9.8 = 490 N; m=? a=1.4 ms-2
F 490
M = = 350 kg.
A 1.4
A car initially at rest, picks up a velocity of 72 kmhr-1 in 20s. If the mass of the car is 1000 kg., calculate (I) Force developed by its engine (ii) Distance covered by the car.
Ans: u=0; v=72Kmhr-1 = 20ms-1; t=20 s; m=1000kg; f=? S=?
Applying v = u + at
20 = 0 + a x 20 \a = 1 ms-2
(i) Force developed, F=ma = 1000 kg x 1 ms-2 = 1000 N.
(II) Applying v2 - u2 = 2 As
(20)2 - (0)2 = 2 x 1 x S.
400
S = = 200M.
2
A golfer hits a golf ball at rest. Such that the contact between the ball and golf stick is for 0.1 s. If the golf ball covers a linear distance of 400 m in 2 s, calculate the magnitude of force applied. Mass of golf ball is 50g.
Ans: u=0; t=0.1 s; S = 400m; time for linear distance = 2 s; m=0.05 kg. F=?
\ Distance covered by ball in 2 s = 400 m
\ Distance covered by ball in 1 s = 400 m ¸ 2 = 200 m
\ Final velocity of ball = 200 ms-1
Applying v=u + at
200 = 0 + a x 0.1
\ a = 2000 ms-2.
\ Force acting on ball F = MA = 0.05 X 2000 = 100 N.
A car of mass 800 kg and moving at 54 kmhr-1 is brought to rest over a distance of 15 m. Find the retarding force developed by its bakes.
Ans: m = 800 kg; u = 54 km hr-1 = 15 m-1; v = 0; S=15m; 1=? R.F =?
V2 - U2 = 2As
Applying (0)2 - (15)2 =2a x 15
\ a = -7.5 ms-2.
\ Retardation = 7.5 ms-2.
\ Retarding force= mass x Retardation
\ = 0.1 kg x 600ms-2= 6000 N.
10 A cricket player holds a cricket ball of mass 100g by moving his hands backward
By 0.75 m. If the initial velocity of ball is 108 kmhr-1. Calculate the retarding force applied by the player.
Ans: m = 100 g = 0.1 kg; S=0.75 M; u = 108 kmhr-1 = 30 ms-1; Retarding force = ?
Appl;ying v2-u2 = 2As
(0)2 - (30)2 = 2a x 0.75
\ a = -600 ms-2
\ Retardation = 600 ms-2
\ Retarding force = mass x retardation
= 0.1 kg x 600 ms-2 = 60 N.
11. A force of 0.6 gf acts on a glass ball of mass 200 g for 12 s. If initially the ball is at rest, calculate (I) final velocity (ii) distance covered. [g = 1000 cms-2]
Ans: F = 0.6 gf = 0.6 x 1000 = 600 dynes; u=0; m=200 g; t=12s. v = ?
S = ?
Applying F = ma
600 = 200 x a or a = 3 cms-2.
Applying v = u + at
V = 0 + 3 x 12 = 36 cms-1.
1
Applying S = ut + at2
2
1
= 0 x 12 + x 3 x (12)2 = 216 cm
2.
12. A bullet of mass 30 g and moving with velocity x, hits a wooden target with a force of 187.5 N. If the bullet penetrates 80 cm, find the value of x.
Ans: m= 0.03 kg; u = c v = 0; F = -187.5 N; S=0.80M; A=?
F -187.5
\ A = =
x 75
\ a= = 3515.625 ms-2.
1.6
(ii) Force acting on bullet F = ma = 0.025 x 3515.625 = 87.89 N.
15. What extra force should an engine of a car develop, such that its velocity changes from 18 kmhr-1 to 72 kmhr-1, over a distance of 20 m, when mass of car is 900 kg?
Ans: u= 18 Kmhr-1 = 5 ms-1; v= 72 kmhr-1 = 20ms-1; S=20m; a=? m =900kg;
Applying v2 - u2 = 2As
(20)2 - (5)2 = 2 x a x 20
\ 375 = 40a or a = 375 ¸ 40 = 9.375 ms-2.
\Extra force developed F= ma = 900