PART II. Dimensió empírica del projecte BLEARN_AUTONOMY
6. Descripció del projecte, disseny d'instruments i anàlisi de resultats recollits
6.3 Experiència amb la retroalimentació docent
×
× kx
m 2
4000 5
100 2 2 = 50 m/s2 Displacement ( )s =x= 5
100 m
∴ v2=u2 +2as =2 +2×50× 5
100
2
=4+5
= 9
∴ v =3ms−1
Option (b) is correct.
26. Compression in spring Velocity gained by block when the spring is at its natural length will compress the spring.
1 2
1 2
2 2
kx = mv x v m
= k = 3 2
4000 = × ×
×
2 9 5
4000 5 = 45
10000 = 6.7
100 m = 6.7 cm Spring at its natural length.
Block at rest first time due to some reason.
Restoring force brings the spring to its natural length and block attains a velocity of 3ms−1.
KE of the block moves blocks ahead and the spring stretches by 6.7 cm but the block does not return due to same region.
Block is now at rest for the second time.
∴ Distance travelled by block when it comes to rest for the second time
=(5+6.7+6.7+6.7) cm
= 25 cm approx.
Option (b) is correct.
27.
8m× AP=16m×(12L−AP) AP=24L−2AP 3AP =24 L AP= 8L
As the CM of M and S does not change, the CM of the bar shall also not change i.e., the displacement of bar will be zero.
Let x be the displacement of rod.
∴ x×8m+(x+6L)48m+(x+12L)16m
=(6L×48m)+(8L×24m) i.e., x = 0 m
i.e., no displacement of bar.
Option (d) is correct.
28. vCM of M and S = + −
+ =
8 2 16
8 16 0
m v m v
m m
( ) ( )
There CM of M and S will not change while they move i.e.,the point P (where they meet) is at the edge of the table supporting the end B.
Option (b) is correct.
5cm
F 6.7 cm
6.7 cm
6.7 cm F
3 m/s
48 m M = 8 m1
2v (M + s) M = 16 m2
2v
A B
6 L
12 L P
29. When the spider eats up the moth and travels towards A with velocity v
2 relative to rod.
[vR= velocity (absolute) of rod]
⇒ v v
R= − 6 Option (c) is correct.
30. Time taken by spider to reach point A starting from point B
=4 + 8 31. Form CM not to shift
⇒ (x′ +8L)24m+(x′ +6L)48m
=64×48 m i.e., x L
′ = −8 3
Option (a) is correct.
More than One Cor rect Op tions 1. Along vertical : 2
2 45
Along horizontal : mv mv Squaring and adding Eq. (i)and (ii), 8
Thus, the divergence angle between the particles will be less than π
2. Option (b) is correct.
Initial KE =1
As Final KE < Initial KE Collision is inelastic.
Option (d) is correct.
2. v m m
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
| 197
48 m
T mv Option (a) is correct.
Velocity of block
v m Option (c) is correct.
Maximum height attained by pendulum bob
Option (d) is correct.
3. vcosφ =ucosθ ⇒ v Option (b) is correct.
Change in momentum of particle
= −( mvsin )φ − +( musin )θ
∴ Impulse delivered by floor to the particle = mvsinφ +musinθ Option (d) is correct.
u 1−(1−e2) sin θ2
Option (c) is correct.
cos2θ+ e2sin2θ Option (d) is correct.
4. →u=(3^i+2^j)ms−1
Impulse received by particle of mass m
= −m→u+m→v
= −m(3^i+2^j)+m(−2^i+^j)
= −m (5 i^+^j unit) Option (b) is correct.
Impulse received by particle of mass M
= − (impulse received by particle of
mass m)
=m (5 i^+^j)
Option (d) is correct.
v
5. T=m a1
Option (b) is correct.
( )
Option (c) is correct.
6. As the block comes down, the CM of the system will also come down i.e., it does not remain stationary. Option (d) is correct.
As no force acts along horizontal direction, the momentum of the system will remain conserved along horizontal direction.
Option (c) is correct.
7. Velocity of B after collision :
v e Option (b) is correct.
Velocity of A after collision
v e Loss of KE during collision
=1 − ′ + ′
Option (c) is correct.
8. As the mass of the system keeps on decreasing momentum of the system does not remain constant.
Thrust force is developed on the rocket due to Newton’s 3rd law of motion.
Option (b) is correct.
As, a dv
The value of a will remain constant if vi and −dm
dt are constant.
Option (c) is correct.
Fnet =Ft (Thrust force due to gas ejection)
− W (weight of rocket) a F
= mnet
Thus, Newton’s 2nd law is applied.
Option (d) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
| 199
m1
Match the Columns
1. If x0 is the compression made in the spring, the restoring force on B will decrease from kx0 to zero as the spring regains its original length. Thus, the acceleration of B will also decrease from kx
mB
0 to zero.
So, the aCM will also decrease from kx
mA mB
0
+ to zero.
∴ (a) → (r)
When spring is released after compressing it, the restoring on B will accelerate it towards right while the reaction force on A will apply a force on the wall which in turn will apply equal and opposite force on A and consequently A will travel towards right. As both travel towards right the velocity of CM will be maximum in the beginning.
After this A will start compressing the spring and at a certain instant when the spring is compressed to maximum value both the blocks will travel towards right with a constant velocity and then the velocity of CM will become constant.
∴ (b) → (q)
As the blocks will never move along y-axis, the y-component of the CM of the two blocks will not change.
∴ (d) → (p)
As the two blocks will keep on moving towards right (surface below being smooth) the x-coordinate of the CM of the blocks will keep on increasing.
∴ (c) → (s)
2. Initial a m g m g
m m
CM= + + +
+
( ) ( )
= + g
= + 10 SI unit
∴ (a) → (q)
Initial v m m
m m
CM= − + ×
+
( 20) 0
= − 10
∴ |vCM|= 10 SI unit
∴ (b) → (q)
For the time taken by the first particle to return to ground
s=ut+1at 2
2
0= −( 20) t+5t2
⇒ t = 4 s
Now, as the collision of the first particle with the ground is perfectly inelastic, the first particle will remain on ground at rest.
Now, let us find the position of 2nd particle at t = 5 s
s =( )0 5+1( ) 2 10 52 = 125 m
The particle (2nd) will still be in space moving downwards.
a m m g
m m
CM= ⋅ + ⋅ + 0
= g =
2 5 (SI unit)
∴ (c) → (p)
Velocity of 2nd particle at t = 5 s v = +0 10×5
= 50 ms−1 B
A
mB mA
k
B A
x0
kx0
kx0
20 ms–1
m
1st particle +
180 m
2nd particle u = 0
∴ At t = 5 s
v m m
m m
CM= ⋅ + ⋅ +
0 50
= 25 (SI unit)
∴ (d) → (s)
3. Initial KE of block B = 4 J
∴ 1
2×0.5×u2=4
⇒ u = 4 ms−1
∴ Initial momentum of B =0.5×4 =2kg ms−1
∴ (a) → (r)
Initial momentum pCM= pA + pB = +0 2 =2kg ms−1
∴ (b) → (r)
Velocity given to block B will compress the spring and this will gradually increase the velocity of A. When the spring gets compressed to its maximum both the blocks will have the same velocities i.e., same momentum as both have same mass.
pA= pB
(at maximum compression of the spring) But, pA+ pB= initial momentum of B.
∴ pA+ pA= 2
i.e., pA = 1 kgms−1
∴ (c) → (q)
After the maximum compression in the spring, the spring will gradually expand but now the velocity of block A will increase and that of B will decrease and when the spring attains maximum expansion the velocity of B will be zero and so will be its momentum.
∴ (d) → (p)
4. If collision is elastic, the two blocks will interchange there velocities (mass of both balls being equal).
Thus, velocity of A after collision = v
∴ (a) → (r)
If collision is perfectly inelastic, the two balls will move together (with velocities V).
∴ mv=(m+m V)
⇒ V v
=2
∴ (b) → (s)
If collision is inelastic with e =1 2,
v e
1 v2
1
′ = 2+
⋅
= +
⋅ 1 1
2
2 v [Q v2= (given)]v
=3 4v
∴ (c) → (p)
If collision is inelastic with e =1 4,
v1 v v
1 1 4 2
5
′ = 8
+
=
∴ (d) → (q).
5. If A moves x towards right
Let plank (along with B) move by x′ to the right.
∴ x× + ′x +
+ + =
30 60 30
30 60( 30 ) 0
( )
i.e., x x
′ = − 3 = x
3, towards left.
∴ (a) → (r)
If B moves x towards left
Let plank (along with A) move x′ to the left
∴ x⋅ + ′x +
+ + =
60 30 30
60 30( 30 ) 0
( )
i.e., x′ = −x
= x, towards right
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
| 201
B A
m = 0.5 kg m
C 30 kg B A
50 kg 60 kg
Smooth