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CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
EXERCISES Q-11 to Q-20
11. Find the center of mass of a uniform plate having semicircular inner and outer boundaries of radii R1 and R2 (Figure 9-E5).
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| Figure for Q-11 |
ANSWER: Take the center of the semicircle as the origin. Due to symmetry about the y-axis, the CoM will lie on the y-axis (x coordinate zero). Let us calculate the y coordinate of the CoM. The given lunar part can be considered as a semicircle of radius R1 taken out of larger semicircle of radius R2. CoM of smaller semicircular plate = 4R1/3𝝅 from the center and CoM of larger semicircular plate = 4R2/3𝝅.
Hence y coordinate of the CoM
= (1/M)[(m𝝅R2² /2)(4R2/3𝝅)-(m𝝅R1² /2)(4R1/3𝝅)]
{where m is the mass per unit area and M the total mass of the object}
= (2m/3M)[(R2³-R1³]
= {2m/3m𝝅(R2²-R1²)/2}*[(R2³-R1³]
{mass of the plate m*𝝅(R2²-R1²)/2 = m𝝅(R2²-R1²)/2}
=4(R2³-R1³)/3𝝅(R2²-R1²)
=4(R2-R1)(R2²+R2R1+R1²)/3 𝝅(R2-R1)(R2+R1)
= 4(R2²+R2R1+R1²)/3 𝝅(R2+R1) above the center.
Hence y coordinate of the CoM
= (1/M)[(m𝝅R2² /2)(4R2/3𝝅)-(m𝝅R1² /2)(4R1/3𝝅)]
{where m is the mass per unit area and M the total mass of the object}
= (2m/3M)[(R2³-R1³]
= {2m/3m𝝅(R2²-R1²)/2}*[(R2³-R1³]
{mass of the plate m*𝝅(R2²-R1²)/2 = m𝝅(R2²-R1²)/2}
=4(R2³-R1³)/3𝝅(R2²-R1²)
=4(R2-R1)(R2²+R2R1+R1²)/3 𝝅(R2-R1)(R2+R1)
= 4(R2²+R2R1+R1²)/3 𝝅(R2+R1) above the center.
12. Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in the water. To discuss a mechanics problem. They come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the progress?
50 kg ..... at zero m (Mr Verma)
40 kg...... at 2 m (Boat)
60 kg ..... at 4 m (Mr Mathur)
Total mass = 50+40+60 =150 kg.
If the distance of CoM of the system be at distance X from the origin,
X = (1/150){50*0+40*2+60*4}
=(80+240)/150
=320/150
=2.13 m
When both the persons come at the middle of the boat, now masses and distances change.
50+60 = 110 kg ...at 2 m (both persons)
40 kg ......at 2 m (Boat)
Now distance of CoM from the same end X'
= (1/150){110*2+40*2}
= 300/150 = 2 m
The above distances of CoM are from one end of the boat. Since there is no external force on the system, the CoM will remain at the same place with respect to the ground. In fact, the boat moves by the distance 2.13 - 2.00 m = 0.13 m = 13 cm.
13. A cart of mass M is at rest on a frictionless horizontal surface and a pendulum bob of mass m hangs from the roof of the cart (Figure 9-E6). The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is L. Find the displacement of the cart during this process.
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| Figure for Q-13 |
Let us take the end of the cart towards string as a reference line, the distance of CoM of bob from this line as x and distance of CoM of the cart from the line as y.
Initial CoM of the system w.r.t. end line X =(1/m+M){mx+My}
Final CoM of the system w.r.t. end line X' = (1/m+M){m(x+L)+My}
Hence movement of the cart =X'-X
=(1/m+M){mx+mL+My-mx-My}
=(1/m+M){mL}
=mL/(m+M)
Hence the cart moves through mL/(m+M) distance towards left.
14. The balloon, the light rope and the monkey shown in the figure (9-E7) are at rest in the air. If the monkey reaches the top of the rope, by what distance does the balloon descend? Mass of the balloon = M, mass of the monkey = m and the length of the rope ascended by the monkey = L.
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| Figure for Q-14 |
the distance of CoM of the monkey = y,
then CoM of the system from the reference point
= X = (1/m+M){Mx+my}
When the monkey reaches the top of the rope, the new distance of the CoM of the monkey =y-L.
Now the distance of the CoM of the system from the reference point = X' = (1/m+M){Mx+m(y-L)}
The relative change in the CoM from the reference point=X-X'
= (1/m+M){Mx+my}- (1/m+M){Mx+m(y-L)}
=(1/m+M){mL} = mL/(m+M)
Since there is no external force on the system, the CoM of the system will remain stationary and the relative change in CoM of the system is in fact the distace by which the balloon descends i.e. = mL/(m+M)
15. Find the ratio of the linear momenta of two particles of masses 1.0 kg and 4.0 kg if their kinetic energies are equal.
K.E. of the first block = ½mv² =½*1.0*v² = v²/2
K.E. of the second block = ½*4.0*v'² =2v'²
Since given that both are equal, hence,
v²/2 = 2v'²
→v² = 4v'²
→v²/v'² =4
→v/v' = 2
Momentum of the first particle = mv =1.0*v = v
Momentum of the second particle = 4.0*v' = 4v'
So ratio of their momenta = v/4v' = 2/4 = 1/2 =1:2
16. A uranium-238 nucleus, initially at rest, emits an alfa particle with a speed of 1.4 x 107 m/s. Calculate the recoil speed of the residual nucleus thorium-234. Assume that the mass of a nucleus is proportional to the mass number.
the mass of residual thorium-234 = 234m kg and the mass of the alfa particle =4m kg, where m is the constant of proportionality.
Since the parent atom is at rest initially, its momentum is zero. Let the recoil speed of the thorium atom be v, then it's momentum
= 234mv kg-m/s
Momentum of the alfa particle
=4m*1.4 x 107 = m5.6 x 107 kg-m/s
Combined momentum of the alfa particle and the residual nucleus
=234mv+m5.6 x 107 kg-m/s
Since there is no external force on the system total momentum will be conserved, here equal to zero.
So, 234mv+m5.6 x 107 = 0
v = -(5.6/234)*107 m/s =- (2.4x10-2)*107 = -2.4x105 m/s
The negative sign shows that the recoil speed is in opposite direction to v, the speed of alfa particle.
17. A man of mass 50 kg starts moving on the earth and acquires a speed of 1.8 m/s. With what speed does the earth recoil? Mass of the earth = 6 x 1024 kg.
(50 kg*1.8 m/s)/6 x 1024 kg
=90/6 x 1024 m/s
=15x10-24 m/s
=1.5x10-23 m/s
18. A neutron initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of 1.4x10-26
kg-m/s and the antineutrino 6.4x10-27 kg-m/s. Find the recoil speed of the proton (a) if the electron and the antineutrino are ejected along the same direction and (b) if they are ejected along perpendicular directions. Mass of the proton = 1.67x10-27 kg.
ANSWER: Since the neutron is initially at rest, total momenta of all the particle will be zero even after the decay because there is no external force.
(a) Sum of momenta of ejected electron and antineutrino
= 1.4x10-26 + 6.4x10-27 kg-m/s
= 14x10-27 + 6.4x10-27 kg-m/s
= 20.4x10-27 kg-m/s
So the recoil speed of the proton
= 20.4x10-27 / 1.67x10-27 m/s
= 12.2 m/s
(b) Let the recoil speed (v) of the proton be along the x-axis. Magnitude of its momentum
= 1.67x10-27*v kg-m/s
The resultant of the momentum of the electron and the antineutrino will be along the x-axis but opposite to the proton. Since both are perpendicular hence the magnitude of the resultant momentum
=√{(14)²+(6.4)²}x10-27 kg-m/s
=√236.96 x10-27 kg-m/s
=15.39x10-27 kg-m/s
Hence the velocity of he proton
v = 15.39x10-27 /1.67x10-27 m/s = 9.22 m/s
Alternately
If the electron is ejected along a direction making an angle θ from the x-axis, the direction of the antineutrino will make angle =90°-θ from the x-axis. See the figure below:-
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| Figure for Q-18 |
Component of the momentum of the electron along the x-axis
= 1.4x10-26.cosθ kg-m/s
and along y-axis = 1.4x10-26.sinθ kg-m/s
Component of the momentum of the antineutrino along the x-axis
= 6.4x10-27.cso(90°-θ) kg-m/s
= 6.4x10-27.sinθ kg-m/s
and along the y-axis = 6.4x10-27.cosθ kg-m/s
Since the sum of the components of the final momenta of the particles along y-axis will also be zero, hence magnitude will be equal and direction opposite. Equating them,
1.4x10-26.sinθ = 6.4x10-27.cosθ
→tanθ = 6.4x10-1/1.4 = 0.46
We can calculate the values of cosθ and sinθ.
cosθ=1/√(1+tan²θ) = 1/√1.21 =0.91
and sinθ = cosθ.tanθ =0.91*0.46 = 0.42
Now equating the moments along x-axis,
1.67x10-27*v = 1.4x10-26.cosθ + 6.4x10-27.sinθ
→ 1.67x10-27*v = 1.4x10-26*0.91 + 6.4x10-27*0.42
→ v = 14x0.91/1.67 + 6.4x0.42/1.67 = 7.63+1.61 =9.24 m/s
19. A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically (Figure 9-E8). When at height h from the ground he notices that the ground below him is pretty hard, but there is a pond at a horizontal distance x from the line of fall. In order to save himself, he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land?
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| Figure for Q-19 |
t = time in falling through H - time in falling through (H-h)
=√(2H/g)-√{2(H-h)/g}
In this time the man covers a horizontal distance of x, so the horizontal velocity of the man
V=x/t = x/[√(2H/g)-√{2(H-h)/g}]
If the horizontal velocity of the bag is v,
then, mv = MV
v = MV/m
=Mx/[√(2H/g)-√{2(H-h)/g}].m
=Mx√g/m[√(2H)-√{2(H-h)}]
The distance traveled by the bag with this velocity in time t
=vt
=Mx√g/m[√(2H)-√{2(H-h)}]*[√(2H/g)-√{2(H-h)/g}]
=Mx/m away from the line of fall opposite to the direction of the pond.
20. A ball of mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence 45°. The ball is reflected by the plane at equal angle of reflection with the same speed. Calculate (a) the magnitude of the change in momentum of the ball (b) the change in the magnitude of the momentum of the ball.
v = 2.0*cos45° = 2/√2 m/s
After the strike, the component of the speed perpendicular to the plane is same but opposite in direction i.e. -v.
So the change in speed = v-(-v) = 2v
(a) Hence the magnitude of the change in momentum of the ball
= m*2v
= 0.05 kg*2*2/√2 m/s
= 0.20/√2 kg-m/s
= 0.14 kg-m/s
(b) Since the magnitude of speeds before and after the strike is same hence magnitude of the momenta is also the same. So the change in magnitude of the momentum of the ball is zero.
===<<<O>>>===
Links to the chapter -
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
HC Verma's Concepts of Physics, Chapter-7, Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
HC Verma's Concepts of Physics, Chapter-6, Friction
Click here for → Friction OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → Friction - OBJECTIVE-II
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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HC Verma's Concepts of Physics, Chapter-4, The Forces
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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