Motion in a Plane Practice Questions - DPP for JEE

 Page 1


1. (b) ,
 ? AB sin ? = v3 AB cos ?
or, tan ? = v3,   ? ? = 60º
2. (b)
? 
3. (c) Clearly   
= 
4. (c) Using the law of vector addition,  is as shown in the fig.
Page 2


1. (b) ,
 ? AB sin ? = v3 AB cos ?
or, tan ? = v3,   ? ? = 60º
2. (b)
? 
3. (c) Clearly   
= 
4. (c) Using the law of vector addition,  is as shown in the fig.
5. (b)
The velocity of first particle, v
1
 = v
The velocity of second particle, v
2
 = at
Relative velocity, 
or 
For least value of relative velocity, 
or  
or  0 + a
2
 × 2t – 2vacos a = 0
or  
6. (d)
R = 10 cos 30° t –  g sin 30° t
2
=  = 
7. (c) = 1 cm/s. Centripetal acceleration is directed towards the
centre. Its magnitude = 1. Unit vector at the mid point on the path
between P and Q is  .
8. (b) Two vectors are
 = 
Page 3


1. (b) ,
 ? AB sin ? = v3 AB cos ?
or, tan ? = v3,   ? ? = 60º
2. (b)
? 
3. (c) Clearly   
= 
4. (c) Using the law of vector addition,  is as shown in the fig.
5. (b)
The velocity of first particle, v
1
 = v
The velocity of second particle, v
2
 = at
Relative velocity, 
or 
For least value of relative velocity, 
or  
or  0 + a
2
 × 2t – 2vacos a = 0
or  
6. (d)
R = 10 cos 30° t –  g sin 30° t
2
=  = 
7. (c) = 1 cm/s. Centripetal acceleration is directed towards the
centre. Its magnitude = 1. Unit vector at the mid point on the path
between P and Q is  .
8. (b) Two vectors are
 = 
 = 
For two vectors  and  to be orthogonal  A.B = 0
 = 0 = cos ?t.cos + sin ?t.sin
= cos
So, ?   t = 
9. (a) due North;
 due West. Angle between  and 
 due South
? Change in velocity
The direction of this change in velocity is in South-West.
10. (b)
Comparing with , we get
v
x
 = 6ms
–1
 and v
y
 = 8 ms
–1
Also,  = 36 + 64 = 100
or v = 10 ms
–1
Page 4


1. (b) ,
 ? AB sin ? = v3 AB cos ?
or, tan ? = v3,   ? ? = 60º
2. (b)
? 
3. (c) Clearly   
= 
4. (c) Using the law of vector addition,  is as shown in the fig.
5. (b)
The velocity of first particle, v
1
 = v
The velocity of second particle, v
2
 = at
Relative velocity, 
or 
For least value of relative velocity, 
or  
or  0 + a
2
 × 2t – 2vacos a = 0
or  
6. (d)
R = 10 cos 30° t –  g sin 30° t
2
=  = 
7. (c) = 1 cm/s. Centripetal acceleration is directed towards the
centre. Its magnitude = 1. Unit vector at the mid point on the path
between P and Q is  .
8. (b) Two vectors are
 = 
 = 
For two vectors  and  to be orthogonal  A.B = 0
 = 0 = cos ?t.cos + sin ?t.sin
= cos
So, ?   t = 
9. (a) due North;
 due West. Angle between  and 
 due South
? Change in velocity
The direction of this change in velocity is in South-West.
10. (b)
Comparing with , we get
v
x
 = 6ms
–1
 and v
y
 = 8 ms
–1
Also,  = 36 + 64 = 100
or v = 10 ms
–1
 = 9.6 m
11. (b) y = bx
2
Differentiating w.r.t to t an both sides, we get
v
y
 = 2bxv
x
Again differentiating w.r.t to t on both sides we get
[ = 0, because the particle has constant acceleration along y-
direction]
Now, 
v
x
 = 
12. (c)
Since  and  are perpendicular
? 
    A
2
 = B
2 
 
13. (a) Total area around fountain
Page 5


1. (b) ,
 ? AB sin ? = v3 AB cos ?
or, tan ? = v3,   ? ? = 60º
2. (b)
? 
3. (c) Clearly   
= 
4. (c) Using the law of vector addition,  is as shown in the fig.
5. (b)
The velocity of first particle, v
1
 = v
The velocity of second particle, v
2
 = at
Relative velocity, 
or 
For least value of relative velocity, 
or  
or  0 + a
2
 × 2t – 2vacos a = 0
or  
6. (d)
R = 10 cos 30° t –  g sin 30° t
2
=  = 
7. (c) = 1 cm/s. Centripetal acceleration is directed towards the
centre. Its magnitude = 1. Unit vector at the mid point on the path
between P and Q is  .
8. (b) Two vectors are
 = 
 = 
For two vectors  and  to be orthogonal  A.B = 0
 = 0 = cos ?t.cos + sin ?t.sin
= cos
So, ?   t = 
9. (a) due North;
 due West. Angle between  and 
 due South
? Change in velocity
The direction of this change in velocity is in South-West.
10. (b)
Comparing with , we get
v
x
 = 6ms
–1
 and v
y
 = 8 ms
–1
Also,  = 36 + 64 = 100
or v = 10 ms
–1
 = 9.6 m
11. (b) y = bx
2
Differentiating w.r.t to t an both sides, we get
v
y
 = 2bxv
x
Again differentiating w.r.t to t on both sides we get
[ = 0, because the particle has constant acceleration along y-
direction]
Now, 
v
x
 = 
12. (c)
Since  and  are perpendicular
? 
    A
2
 = B
2 
 
13. (a) Total area around fountain
Where R
max
 = 
? 
14. (a)
and  
15. (a) Distance covered in one circular loop = 2pr
= 2 × 3.14 × 100 = 628 m
Speed 
Displacement in one circular loop = 0
Velocity 
16. (a)
Now