Laws of Motion Practice Questions - DPP for JEE

 Page 1


1. (d) Here m = 0.5 kg ; u = – 10 m/s;
t = 1/50 s ; v = + 15 ms
–1
Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N
2. (b) From figure,
Acceleration a = Ra …(i)
and mg – T = ma …(ii)
From equation (i) and (ii)
T × R = mR
2
a = mR
2
or T = ma
? mg – ma  = ma
?
3. (a) Limiting friction between block and slab
= µ
s
m
A
g = 0.6 ×10 × 9.8 = 58.8 N
But applied force on block A is 100 N. So the block will slip over a slab.
Now kinetic friction works between block and slab
F
k
 = µ
k
m
A
g = 0.4 × 10× 9.8 = 39.2 N
This kinetic friction helps  to move the slab
? Accleration of slab 
Page 2


1. (d) Here m = 0.5 kg ; u = – 10 m/s;
t = 1/50 s ; v = + 15 ms
–1
Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N
2. (b) From figure,
Acceleration a = Ra …(i)
and mg – T = ma …(ii)
From equation (i) and (ii)
T × R = mR
2
a = mR
2
or T = ma
? mg – ma  = ma
?
3. (a) Limiting friction between block and slab
= µ
s
m
A
g = 0.6 ×10 × 9.8 = 58.8 N
But applied force on block A is 100 N. So the block will slip over a slab.
Now kinetic friction works between block and slab
F
k
 = µ
k
m
A
g = 0.4 × 10× 9.8 = 39.2 N
This kinetic friction helps  to move the slab
? Accleration of slab 
4. (b) Thrust on the satellite,
Acceleration 
5. (c) Let T be the tension in the branch of a tree when monkey is
descending with acceleration a. Then mg – T = ma; and T = 75%
of weight of monkey
 or .
6. (a) Coefficient of static friction,
µ
s
 = tan 30° =  = 0.577 ? 0.6
S = ut + 
4 = a(4)
2
 ? a =  = 0.5
[ s = 4m and t = 4s given]
a = gsin? – µ
k
(g) cos?
? µ
k 
=  = 0.5
7. (d) According to law of conservation of momentum the third piece has
momentum
kg ms
–1
Impulse = Average force × time
? Average force 
 
8. (d) Given F = 600 – (2 × 10
5
 t)
The force is zero at time t, given by
Page 3


1. (d) Here m = 0.5 kg ; u = – 10 m/s;
t = 1/50 s ; v = + 15 ms
–1
Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N
2. (b) From figure,
Acceleration a = Ra …(i)
and mg – T = ma …(ii)
From equation (i) and (ii)
T × R = mR
2
a = mR
2
or T = ma
? mg – ma  = ma
?
3. (a) Limiting friction between block and slab
= µ
s
m
A
g = 0.6 ×10 × 9.8 = 58.8 N
But applied force on block A is 100 N. So the block will slip over a slab.
Now kinetic friction works between block and slab
F
k
 = µ
k
m
A
g = 0.4 × 10× 9.8 = 39.2 N
This kinetic friction helps  to move the slab
? Accleration of slab 
4. (b) Thrust on the satellite,
Acceleration 
5. (c) Let T be the tension in the branch of a tree when monkey is
descending with acceleration a. Then mg – T = ma; and T = 75%
of weight of monkey
 or .
6. (a) Coefficient of static friction,
µ
s
 = tan 30° =  = 0.577 ? 0.6
S = ut + 
4 = a(4)
2
 ? a =  = 0.5
[ s = 4m and t = 4s given]
a = gsin? – µ
k
(g) cos?
? µ
k 
=  = 0.5
7. (d) According to law of conservation of momentum the third piece has
momentum
kg ms
–1
Impulse = Average force × time
? Average force 
 
8. (d) Given F = 600 – (2 × 10
5
 t)
The force is zero at time t, given by
0 = 600 – 2 × 10
5
 t
 seconds
?  Impulse
9. (d) As shown in the figure, the three forces are represented by the
sides of a triangle taken in the same order. Therefore the resultant
force is zero.
Therefore acceleration is also zero i.e., velocity remains
unchanged.
10. (b) From the F.B.D.
N = mg cos
F = ma = mg sin – µN
?   a = g(sin ? – µ cos ?)
Now using, v
2
 – u
2
 = 2as
or,  
( l = length of incline)
or, v = 
11. (a) As the ball, m = 10 g = 0.01 kg rebounds after striking the wall
? Change in momentum = mv – (–mv) = 2 mv
Inpulse = Change in momentum = 2mv
Page 4


1. (d) Here m = 0.5 kg ; u = – 10 m/s;
t = 1/50 s ; v = + 15 ms
–1
Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N
2. (b) From figure,
Acceleration a = Ra …(i)
and mg – T = ma …(ii)
From equation (i) and (ii)
T × R = mR
2
a = mR
2
or T = ma
? mg – ma  = ma
?
3. (a) Limiting friction between block and slab
= µ
s
m
A
g = 0.6 ×10 × 9.8 = 58.8 N
But applied force on block A is 100 N. So the block will slip over a slab.
Now kinetic friction works between block and slab
F
k
 = µ
k
m
A
g = 0.4 × 10× 9.8 = 39.2 N
This kinetic friction helps  to move the slab
? Accleration of slab 
4. (b) Thrust on the satellite,
Acceleration 
5. (c) Let T be the tension in the branch of a tree when monkey is
descending with acceleration a. Then mg – T = ma; and T = 75%
of weight of monkey
 or .
6. (a) Coefficient of static friction,
µ
s
 = tan 30° =  = 0.577 ? 0.6
S = ut + 
4 = a(4)
2
 ? a =  = 0.5
[ s = 4m and t = 4s given]
a = gsin? – µ
k
(g) cos?
? µ
k 
=  = 0.5
7. (d) According to law of conservation of momentum the third piece has
momentum
kg ms
–1
Impulse = Average force × time
? Average force 
 
8. (d) Given F = 600 – (2 × 10
5
 t)
The force is zero at time t, given by
0 = 600 – 2 × 10
5
 t
 seconds
?  Impulse
9. (d) As shown in the figure, the three forces are represented by the
sides of a triangle taken in the same order. Therefore the resultant
force is zero.
Therefore acceleration is also zero i.e., velocity remains
unchanged.
10. (b) From the F.B.D.
N = mg cos
F = ma = mg sin – µN
?   a = g(sin ? – µ cos ?)
Now using, v
2
 – u
2
 = 2as
or,  
( l = length of incline)
or, v = 
11. (a) As the ball, m = 10 g = 0.01 kg rebounds after striking the wall
? Change in momentum = mv – (–mv) = 2 mv
Inpulse = Change in momentum = 2mv
? ? = 
12. (a) At limiting equilibrium, µ = tan?
tan? = µ = (from question)
Q Coefficient of friction µ = 0.5
?
? x = + 1
Now, 
13. (d) Writing free body-diagrams for m & M,
we get   T = ma  and F – T = Ma
where T is force due to spring
 F – ma = Ma   or, F = Ma + ma
  
Now, force acting on the block of mass m is
Page 5


1. (d) Here m = 0.5 kg ; u = – 10 m/s;
t = 1/50 s ; v = + 15 ms
–1
Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N
2. (b) From figure,
Acceleration a = Ra …(i)
and mg – T = ma …(ii)
From equation (i) and (ii)
T × R = mR
2
a = mR
2
or T = ma
? mg – ma  = ma
?
3. (a) Limiting friction between block and slab
= µ
s
m
A
g = 0.6 ×10 × 9.8 = 58.8 N
But applied force on block A is 100 N. So the block will slip over a slab.
Now kinetic friction works between block and slab
F
k
 = µ
k
m
A
g = 0.4 × 10× 9.8 = 39.2 N
This kinetic friction helps  to move the slab
? Accleration of slab 
4. (b) Thrust on the satellite,
Acceleration 
5. (c) Let T be the tension in the branch of a tree when monkey is
descending with acceleration a. Then mg – T = ma; and T = 75%
of weight of monkey
 or .
6. (a) Coefficient of static friction,
µ
s
 = tan 30° =  = 0.577 ? 0.6
S = ut + 
4 = a(4)
2
 ? a =  = 0.5
[ s = 4m and t = 4s given]
a = gsin? – µ
k
(g) cos?
? µ
k 
=  = 0.5
7. (d) According to law of conservation of momentum the third piece has
momentum
kg ms
–1
Impulse = Average force × time
? Average force 
 
8. (d) Given F = 600 – (2 × 10
5
 t)
The force is zero at time t, given by
0 = 600 – 2 × 10
5
 t
 seconds
?  Impulse
9. (d) As shown in the figure, the three forces are represented by the
sides of a triangle taken in the same order. Therefore the resultant
force is zero.
Therefore acceleration is also zero i.e., velocity remains
unchanged.
10. (b) From the F.B.D.
N = mg cos
F = ma = mg sin – µN
?   a = g(sin ? – µ cos ?)
Now using, v
2
 – u
2
 = 2as
or,  
( l = length of incline)
or, v = 
11. (a) As the ball, m = 10 g = 0.01 kg rebounds after striking the wall
? Change in momentum = mv – (–mv) = 2 mv
Inpulse = Change in momentum = 2mv
? ? = 
12. (a) At limiting equilibrium, µ = tan?
tan? = µ = (from question)
Q Coefficient of friction µ = 0.5
?
? x = + 1
Now, 
13. (d) Writing free body-diagrams for m & M,
we get   T = ma  and F – T = Ma
where T is force due to spring
 F – ma = Ma   or, F = Ma + ma
  
Now, force acting on the block of mass m is
ma = 
14. (b) The acceleration of mass m is due to the force T cos
? T cos = ma   a = ... (i)
also,F = 2T sin  T = ... (ii)
From (i) and (ii)
a = 
15. (a) The Earth pulls the block by a force Mg. The
block in turn exerts a force Mg on the spring of
spring balance S
1
 which therefore shows a
reading of M kgf.
The spring S
1
 is massless. Therefore it exerts a force of
Mg on the spring of spring balance S
2
 which
shows the reading of M kgf.
16. (b) For the motion of both the blocks
m
1
a = T – µ
k
m
1
g
m
2
g – T = m
2
a