Q: A particle of mass m is moving in the x y -plane such that its velocity at a point ( x , y ) is given as v → = α y x ∧ + 2 x y ∧ where α is a non-zero co

As force is given by, F → = m d v → d t . Given: v → = α y x ∧ + 2 x y ∧ , where velocity along x-axis is v x = α y and velocity along y-axis is v y = 2 x α . Therefore, d v → d t = α d y d t x ∧ + 2 d x d t y ∧ = α v y x ∧ + 2 v x y ∧ = α 2 x α x ∧ + 2 α y y ∧ = 2 α 2 x x ∧ + y y ∧ Therefore, required value of F → = 2 m &

Q: A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3 h from the ground, as shown in the figure. A spherical ball of

Using energy conservation, m g h = 1 2 m u 0 2 ⇒ u 0 = 2 g h u 0 is the horizontal component of the velocity throughout the flight. For vertical component of the velocity(just before collision with the ground), we can use equation of motion v z 2 = 0 2 + 2 - g - 3 h ⇒ v z = 6 g h Angle θ can be written as, tan θ = v z u = 3 ⇒ θ = 60 o Horizontal distance covered just before the collision will be, d = u 0 T = u 0 2 3 h g = 2 g h 2 3 h g = 2 3 h A

Q: A person of height 1 . 6 m is walking away from a lamp post of height 4 m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. I

Given: Speed of person = d x 1 d t = 60 cm s - 1 Also Speed of tip of person's shadow = d x 2 d t . As ∆ A B E and ∆ D C E are similar triangles, therefore we can write 4 x 2 = 1 . 6 x 2 - x 1 ⇒ 4 x 2 - 4 x 1 = 1 . 6 x 2 ⇒ 2 . 4 x 2 = 4 x 1 Differentiate both sides w.r.t. t , we get 2 . 4 d x 2 d t = 4 d x 1 d t ⇒ d x 2 d t = 4 2 . 4 60 = 100 cm s - 1 Now, speed of the tip of the per

Q: A projectile is fired from horizontal ground with speed v and projection angle θ . When the acceleration due to gravity is g , the range of the projectile is d . If at the highest point in

Horizontal range of the projectile is d = v 2 sin 2 θ g Maximum height attained by the projectile is H = v 2 sin 2 θ 2 g So, after entering the new region, time taken by projectile to reach ground t = 2 H g ' = 2 v 2 sin 2 θ 2 g g 0 . 81 = 0 . 94 v sin θ g So, horizontal displacement done by the projectile in new region is x = 0 . 9 v sin θ g × v cos θ = 0 . 9 v 2 sin 2 θ 2 g So, new range is d ' = d 2 + x = v 2 sin 2 θ 2 g +

Question 1

A particle is thrown with a speed v from a point O at an angle with the horizontal plane such that it passes through the point P at a height of 1 m and horizontal distance of 5 m from O, as shown in t

Accepted Answer

The equation of trajectory of a projectile is given by: y = x - g x^2 2 v^2 ^2 The particle passes through point P(5, 1). Substituting x = 5 and y = 1: 1 = 5 - 25 g 2 v^2 ^2 For option (A), substituting = 45^ : 1 = 5 45^ - 25 g 2 v^2 ^2 45^ 1 = 5(1) - 25 g 2 v^2 (1/2) 1 = 5 - 25 g v^2 25 g v^2 = 4 v^2 = 25 g 4 v = 5 g 2 ms^ -1 Thus, option (A) is correct. For option (B), the horizontal distance to the maximum height is x_m = R 2 = v^2 (2 ) 2g . For = 45^ and v^2 = 25g 4 : x_m = ( 25g 4 ) 90^ 2g

Question 2

A projectile of mass 200 g is launched in a viscous medium at an angle 60^ with the horizontal, with an initial velocity of 270 ~m / s . It experiences a viscous drag force F =-c v where the drag coef

Accepted Answer

F _ net = m ~d v dt mg + F = md v dt mg - C v = md v dt Horizontal direction - Cv _ x = mdv _ x dt aligned & - C m _0^t d t= _ v_ 0 x ^ v_x d v_x v_x \ & - t 2 = n v_x v_ 0 x \ & d x d t =v_x=v_ 0 x e^ -t / 2 \ & _0^ S_x d x=v_ 0 x _0^t e^ -t / 2 d t \ & S_x=2 v_ 0 x (1-e^ -t / 2 ) \ & at t=2 aligned aligned & S _ x =2 270 60^ [1- 1 e ] \ & S _ x =270 (1- 1 2.7 ) \ & = 270 2.7 (1.7) \ & =170 ~m \ & ~S _ x =170 ~m aligned

Question 3

A particle of mass m  is moving in the x y -plane such that its velocity at a point ( x , &#160; y ) is given as v &#8594; = &#945; y x &#8743; + 2 x y &#8743;  where &#945; is a non-zero co

Accepted Answer

As force is given by,  F &#8594; = m d v &#8594; d t . Given:  v &#8594; = &#945; y x &#8743; + 2 x y &#8743; , where velocity along x-axis is  v x = &#945; y  and velocity along y-axis is  v y = 2 x &#945; . Therefore, d v &#8594; d t = &#945; d y d t x &#8743; + 2 d x d t y &#8743; = &#945; v y x &#8743; + 2 v x y &#8743; = &#945; 2 x &#945; x &#8743; + 2 &#945; y y &#8743; = 2 &#945; 2 x x &#8743; + y y &#8743; Therefore, required value of  F &#8594; = 2 m &

Question 4

A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3 h from the ground, as shown in the figure. A spherical ball of

Accepted Answer

Using energy conservation, m g h = 1 2 m u 0 2 &#8658; u 0 = 2 g h u 0  is the horizontal component of the velocity throughout the flight. For vertical component of the velocity(just before collision with the ground), we can use equation of motion v z 2 = 0 2 + 2 - g - 3 h &#8658; v z = 6 g h Angle  &#952;  can be written as, tan &#952; = v z u = 3 &#8658; &#952; = 60 o Horizontal distance covered just before the collision will be, d = u 0 T = u 0 2 3 h g = 2 g h 2 3 h g = 2 3 h A

Question 5

A person of height 1 . 6 &#160; m is walking away from a lamp post of height 4 &#160; m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. I

Accepted Answer

Given: Speed of person  = d x 1 d t = 60 &#160; cm &#160; s - 1 Also Speed of tip of person's shadow  = d x 2 d t .  As  &#8710; A B E  and  &#8710; D C E  are similar triangles, therefore we can write 4 x 2 = 1 . 6 x 2 - x 1 &#8658; 4 x 2 - 4 x 1 = 1 . 6 x 2 &#8658; 2 . 4 x 2 = 4 x 1 Differentiate both sides w.r.t. t , we get 2 . 4 d x 2 d t = 4 d x 1 d t &#8658; d x 2 d t = 4 2 . 4 60 = 100 &#160; cm &#160; s - 1 Now, speed of the tip of the per

Question 6

A projectile is fired from horizontal ground with speed v  and projection angle &#952; . When the acceleration due to gravity is g , the range of the projectile is d . If at the highest point in

Accepted Answer

Horizontal range of the projectile is  d = v 2 sin 2 &#952; g Maximum height attained by the projectile is  H = v 2 sin 2 &#952; 2 g So, after entering the new region, time taken by projectile to reach ground t = 2 H g ' = 2 v 2 sin 2 &#952; 2 g g 0 . 81 = 0 . 94 v sin &#952; g So, horizontal displacement done by the projectile in new region is x = 0 . 9 v sin &#952; g &#215; v cos &#952; = 0 . 9 v 2 sin 2 &#952; 2 g So, new range is  d ' = d 2 + x = v 2 sin 2 &#952; 2 g +

Question 7

A projectile is thrown from a point O on the ground at an angle 45^ from the vertical and with a speed 5 2 ~m / s . The projectile at the highest point of its trajectory splits into two equal parts. O

Accepted Answer

After splitting of projectile in two equal parts one part falls vertically down to the ground.  So, velocity of another part will be according to conservation of linear momentum, m &#215; 5 = m 2 &#215; 0 + m 2 &#215; v &#8658; v = 10 &#160; m &#160; s &#8722; 1 Now time taken by  m 1 , to reach ground,  since after splitting of projectile  m 1  will perform horizontal projectile so, t = 2 h g &#8658; t = 2 u 2 sin 2 &#952; 2 g 2 &#8658; t = 2 &#215; 50 2 &#215; 10

Question 8

A projectile is thrown from a point O on the ground at an angle 45^ from the vertical and with a speed 5 2 ~m / s . The projectile at the highest point of its trajectory splits into two equal parts. O

Accepted Answer

So,   x = R 2 + r   Where, R is the range of oblique projectile and r  is the range of horizontal projectile. x = R 2 + 10 &#215; 2 h g x = u 2 sin 2 &#952; g &#215; 2 + 10 &#215; 0.5 x = 50 20 + 5 x = 2.5 + 5 x = 7.5 &#160; m

Question 9

A uniform electric field, E → = - 400 3 j ^ N C - 1 is applied in a region. A charged particle of mass m carrying positive charge q is projected in this region with an initial speed of 2 10 × 10 6 m s

Accepted Answer

a y = E M = 4 3 × 10 12 m s 2 R = u 2 sin 2 θ a y sin 2 θ = R a y u 2 = 3 2 θ = 30 ° or 60 ° T = 2 u sin 30 ° a y or 2 u sin 60 ° a y = 5 6 μ s or 5 2 μ s

Motion In Two Dimensions — Previous Year Questions & Solutions