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CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
EXERCISES Q-31 to Q-42
31. A ball of mass 0.50 kg moving at a speed of 5.0 m/s collides with another ball of mass 1.0 kg. After the collision, the balls stick together and remain motionless. What was the velocity of the 1.0 kg block before the collision?
ANSWER: Let the velocity of the 1.0 kg block be v m/s before the collision. The total momentum of the two ball-system before the collision
= 0.50*5.0 + 1.0*v =2.5+v
The total momentum of the system after collision = 0
Since there is no external force on the system, the momentum of the system will be conserved. So,
2.5+v = 0
→ v = -2.5 m/s
= 0.50*5.0 + 1.0*v =2.5+v
The total momentum of the system after collision = 0
Since there is no external force on the system, the momentum of the system will be conserved. So,
2.5+v = 0
→ v = -2.5 m/s
So the velocity of the 1.0 kg block before the collision was 2.5 m/s opposite to the direction of motion of the first ball.
32. A 60 kg man skating with a speed of 10.0 m/s collides with a 40 kg skater at rest and they cling to each other. Find the loss of kinetic energy during the collision.
ANSWER: Let the speed of both the skater be v m/s after the collision. Since there is no external force on the two-skater system, their momentum before and after the collision will be the same. Hence,
(60+40)*v = 60*10.0
→ v = 600/100 = 6.0 m/s
Initial Kinetic energy = ½*60*10*10 = 3000 J
Final Kinetic Energy = ½*(60+40)*6*6 = 1800 J
(60+40)*v = 60*10.0
→ v = 600/100 = 6.0 m/s
Initial Kinetic energy = ½*60*10*10 = 3000 J
Final Kinetic Energy = ½*(60+40)*6*6 = 1800 J
Loss of Kinetic energy = 3000 J - 1800 J =1200 J
33. Consider a head-on collision between two particles of masses m1 and m2 The initial speed of the particles are u1 and u2 in the same direction. The collision starts at t = 0 and the particles interact for a time interval Δt. During the collision, the speed of the first particle varies as
v(t) = u1 + (t/Δt)(v1- u1)
Find the speed of the second particle as a function of time during the collision.
ANSWER: Momentum of the system before collision
= m1u1 + m2u2
Let the speed of the second particle be v'(t) during the collision. Momentum of the system during the collision = m1v(t) + m2v'(t) Since in the absence of external force the moment should be conserved, hence equating these two we get,
m1v(t) + m2v'(t) = m1u1 + m2u2
→ v'(t) = (m1/m2)u1 + u2 - (m1/m2).v(t)
→ v'(t) = (m1/m2)u1 + u2 - (m1/m2).{u1 + (t/Δt)(v1- u1)}
→ v'(t) = u2 - (m1/m2).(t/Δt)(v1- u1)
34. A bullet of mass m moving at a speed v hits a ball of mass M kept at rest. A small part having mass m' breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed v1 in the direction of the bullet. Find the velocity of the bullet after the collision.
ANSWER: Momentum of the system before the collision
=mv+M*0 =mv
Let the velocity of the bullet after the collision = v'
Momentum after the collision =(m+m')v'+(M-m')v1 Since there is no external force on the system of the bullet and the ball, the momentum before and after the collision will be the same.
(m+m')v'+(M-m')v1 =mv
→v' = {mv-(M-m')v1}/(m+m') in the initial direction.
35. A ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is three-fourths of the original. Find the coefficient of restitution.
ANSWER: Let the speeds of the balls after the collision be u and u'. From the conservation theorem of momentum, we have
mu+mu' = mv
→u+u'=v ----------------(i)
K.E. of the balls before collision = ½mv²
mu+mu' = mv
→u+u'=v ----------------(i)
K.E. of the balls before collision = ½mv²
K.E. of the balls after collision = ½mu²+½mu'², it is 3/4th of the original. Hence,
½mu²+½mu'² =(3/4)*½mv²
→u²+u'² =3v²/4
Squaring (i),
u²+u'² +2uu'=v²
→2uu' = v² - (u²+u'²) =v²-3v²/4 =¼v²
→uu' = v²/8
Now, (u-u')² = u²+u'²-2uu' =¾v²-¼v² = v²/2
→u-u' =v/√2 = Velocity of separation after the collision.
And velocity of approach before collision = v-0 =v.
The coefficient of restitution =(Velocity of separation after the collision)÷(velocity of approach before collision)
=(v/√2)÷(v)
=1/√2
36. A block of mass 2.0 kg moving at 2.0 m/s collides head-on with another block of equal mass kept at rest. (a) Find the maximum possible loss in kinetic energy due to the collision. (b) If the actual loss in kinetic energy is half of the maximum, find the coefficient of restitution.
ANSWER: (a) Maximum possible loss of K.E. will be in the case of fully inelastic collision when both masses move together at the same speed after the collision. The loss in K.E. for the fully inelastic collision is given by,
= m1m2(v1-v2)²/2(m1+m2)
=2*2(2-0)²/2(2+2)
=16/8
=2 J
(b) Here the actual loss in K.E. =½*2 J = 1J
If u and u' is the speeds after the collision, from moment conservation,
2u+2u' = 2*2
→u+u' = 2
And ½*2*2²-½*2u²-½*2u'² = 1 [Loss in K.E.]
→u²+u'²=4-1=3
= m1m2(v1-v2)²/2(m1+m2)
=2*2(2-0)²/2(2+2)
=16/8
=2 J
(b) Here the actual loss in K.E. =½*2 J = 1J
If u and u' is the speeds after the collision, from moment conservation,
2u+2u' = 2*2
→u+u' = 2
And ½*2*2²-½*2u²-½*2u'² = 1 [Loss in K.E.]
→u²+u'²=4-1=3
If e = coefficient of restitution,
u-u' = e(v1-v2) = e(2-0) =2e
Now, ½[(u+u')²+(u-u')²]= u²+u'²
→½[2²+4e²]=3
→2+2e² =3
→2e²=1
→e²=1/2
→e =1/√2
u-u' = e(v1-v2) = e(2-0) =2e
Now, ½[(u+u')²+(u-u')²]= u²+u'²
→½[2²+4e²]=3
→2+2e² =3
→2e²=1
→e²=1/2
→e =1/√2
37. A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision, what could be the minimum and the maximum value of u?
ANSWER: m=100 g =0.10 kg
If the speeds of particles after the collision = v and v'
Given ½*0.10v²+½0.10v'² = 0.2
→v²+v'² = 0.2*2/0.10 =4
From the momentum conservation,
0.10v+0.10v'=0.1u
→u = v+v'
v²+v'² = ½[(v+v')²+(v-v')²] =½[u²+(v-v')²]
→½[u²+(v-v')²] = 4
→u²+(v-v')²=8
→u²=8-(v-v')²
(i) For the minimum value of u, (v-v')² should be maximum i.e.v-v' should be maximum. Since v-v' is the velocity of separation maximum value of which can be up to the velocity of approach as in the case of a perfectly elastic collision. So here, v-v' = u-0 =u. Putting this value into the equation, we get,
u²=8-u²
→2u²=8
→u²=4
→u=2 m/s
(ii) For maximum value of u, (v-v')² should be minimum i.e. v-v' =0
→v=v' (As in the case of a fully inelastic collision).
So, u²=8-0 =8
→u = 2√2 m/s
38. Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform. (a) Find the speed of A after he rolls the ball for the first time. (b) Find the speed of A after he catches the ball for the first time. (c) Find the speeds of A and B after the ball has made 5 round trips and is held by A. (d) How many times can A roll the ball? (e) Where is the center of mass of the system "A+B+ball" at the end of the nth trip?
ANSWER: The ball has mass m = 4 kg and when in motion a fixed speed v = 5 m/s, hence it has a fixed momentum
= 4 kg x 5 m/s = 20 kg-m/s. (Backward or forward)
(a) Before rolling the ball first time both A and the ball has total linear momentum zero. If the speed of A after rolling the ball is v', then
20 + 40*v' = 0
→ v' =-20/40 = 0.5 m/s opposite to the direction of the ball.
(b) Before catching the ball total momentum of the ball and B is 20 kg-m/s. Let the speed of B after catching the ball be u.
(40+4)*u = 20
→ u = 20/44 = 10/22 m/s. (in the same direction).
Let after throwing the ball speed of B is u'.
40u'-20 = 10/22*44
→u' = 40/40 = 1 m/s
For A,
If the speed after catching the ball is V
(40+4)V = 20 + 40*0.5
→ V = 40/44 = 10/11 m/s
(c) We see that in one round-trip speed of A increases by 10/11 m/s when he catches the ball. So after the fifth round speed of A = 5x10/11 = 50/11 m/s. And the speed of B = 5x 1 m/s = 5 m/s.
= 4 kg x 5 m/s = 20 kg-m/s. (Backward or forward)
(a) Before rolling the ball first time both A and the ball has total linear momentum zero. If the speed of A after rolling the ball is v', then
20 + 40*v' = 0
→ v' =-20/40 = 0.5 m/s opposite to the direction of the ball.
(b) Before catching the ball total momentum of the ball and B is 20 kg-m/s. Let the speed of B after catching the ball be u.
(40+4)*u = 20
→ u = 20/44 = 10/22 m/s. (in the same direction).
Let after throwing the ball speed of B is u'.
40u'-20 = 10/22*44
→u' = 40/40 = 1 m/s
For A,
If the speed after catching the ball is V
(40+4)V = 20 + 40*0.5
→ V = 40/44 = 10/11 m/s
(c) We see that in one round-trip speed of A increases by 10/11 m/s when he catches the ball. So after the fifth round speed of A = 5x10/11 = 50/11 m/s. And the speed of B = 5x 1 m/s = 5 m/s.
(d) After 6th time A rolls the ball if his speed is V' then,
40V' = 44*50/11+20
→V' = 44*50/11*40 + 20/40
→V' =5+ 0.5 =5.5 m/s
Now he can not have the ball because B can not catch it since both B and the ball have the same speed of 5 m/s in the same direction.
So A can roll the ball maximum six times.
(e) Initially when the ball is with A everything is at rest. Let the center of mass be at a distance of X meter from A towards B. Then,
(40+40+4)*X = (40+4)*0+40*d
→X = 40d/84 m =10d/21 m.
Since there is no external force on the system "A+B+ball" the CoM will remain at the same point even after the ball is rolled nth trip. i.e. 10d/21 m from the initial position of A towards B
40V' = 44*50/11+20
→V' = 44*50/11*40 + 20/40
→V' =5+ 0.5 =5.5 m/s
Now he can not have the ball because B can not catch it since both B and the ball have the same speed of 5 m/s in the same direction.
So A can roll the ball maximum six times.
(e) Initially when the ball is with A everything is at rest. Let the center of mass be at a distance of X meter from A towards B. Then,
(40+40+4)*X = (40+4)*0+40*d
→X = 40d/84 m =10d/21 m.
Since there is no external force on the system "A+B+ball" the CoM will remain at the same point even after the ball is rolled nth trip. i.e. 10d/21 m from the initial position of A towards B
39. A ball falls on the ground from a height of 2.0 m and rebounds up to a height of 1.5 m. Find the coefficient of restitution.
ANSWER: Speed just before touching the ground v = √(2gh) =√(4g) =2√g
Speed just after rebound v' = √(2g*1.5) = √(3g)Velocity of separation = v'-0 = v'
Velocity of approach = v-0 = v
Hence coefficient of restitution e = v'/v = √(3g) / 2√g =√3/2
Note: In such cases e =√(h'/h) {Since e =√2gh'/√2gh}
40. In a gamma decay process, the internal energy of a nucleus of mass M decreases, a gamma photon of energy E and linear momentum E/c is emitted and the nucleus recoils. Find the decrease in internal energy.
ANSWER: Let the recoil speed of the nucleus be v. from the Law of conservation of Linear momentum,
Mv = E/c {taking mass of the photon neligible}
Mv = E/c {taking mass of the photon neligible}
v = E/cM
Now the total K.E. of the system = ½M(E/cM)²+E = E²/2Mc²+ ESince this K.E. is gained from the loss of internal energy. Hence in this case decrease in internal energy = E²/2Mc²+ E.
41. A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s (figure 9-E12) towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N/m. Find the maximum compression of the spring.
 |
| Figure for Q-41 |
ANSWER: Let the maximum compression be x. Initial Energy of the system = P.E.+K.E. = 0+½*2*1² = 1 J
At the point of maximum compression, the speed of both the blocks will be the same say v. From the linear momentum conservation,
(2+2)*v = 2*1
→ v = 2/4 = 0.5 m/s
Final Total energy = K.E.+ P.E.
= ½*4*(0.5)²+ ½*100*x²
= 0.50+ 50x²
Since the collision is elastic, initial and final energies will be the same, hence
0.50+ 50x² = 1
x² = 0.50/50 = 0.01
→ x = 0.10 m = 10 cm
42. A bullet of mass 20 g traveling horizontally with a speed of 500 m/s passes through a wooden block of mass 10 kg initially at rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming to rest. Find the friction coefficient between the block and the surface. (Figure 9-E13).
At the point of maximum compression, the speed of both the blocks will be the same say v. From the linear momentum conservation,
(2+2)*v = 2*1
→ v = 2/4 = 0.5 m/s
Final Total energy = K.E.+ P.E.
= ½*4*(0.5)²+ ½*100*x²
= 0.50+ 50x²
Since the collision is elastic, initial and final energies will be the same, hence
0.50+ 50x² = 1
x² = 0.50/50 = 0.01
→ x = 0.10 m = 10 cm
42. A bullet of mass 20 g traveling horizontally with a speed of 500 m/s passes through a wooden block of mass 10 kg initially at rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming to rest. Find the friction coefficient between the block and the surface. (Figure 9-E13).
 |
| Figure for Q-42 |
ANSWER: Assume the velocity of the block v just when the bullet emerges through it. From the Conservation Law of Linear Momentum,
10v + 0.02*100 = 0.02*500 {20g = 20/1000 =0.02 kg}
→ v = 0.02*400/10 = 0.8 m/s
Let the acceleration of the block be = -a, frpm v²=u²+2as we get,
0²=(0.8)²-2a*0.2 {20 cm = 0.2 m}
→a = 0.8*0.8/0.4 = 1.6 m/s²
So friction force = mass*acceleration = 10*(-1.6) = -16 N
10v + 0.02*100 = 0.02*500 {20g = 20/1000 =0.02 kg}
→ v = 0.02*400/10 = 0.8 m/s
Let the acceleration of the block be = -a, frpm v²=u²+2as we get,
0²=(0.8)²-2a*0.2 {20 cm = 0.2 m}
→a = 0.8*0.8/0.4 = 1.6 m/s²
So friction force = mass*acceleration = 10*(-1.6) = -16 N
Negative sign shows that it is against the motion which Friction force always is and its magnitude is 16 N. But from the Friction Law, Friction force = µN
Here N = mg = 10*10 N = 100 N {Taking g = 10 m/s²}
Friction force = µ100 N
Equating both the values of Friction Force we get,
µ100 = 16
→ µ = 16/100 = 0.16
Here N = mg = 10*10 N = 100 N {Taking g = 10 m/s²}
Friction force = µ100 N
Equating both the values of Friction Force we get,
µ100 = 16
→ µ = 16/100 = 0.16
===<<<O>>>===
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CHAPTER- 10 - Rotational Mechanics
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