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Motion question

2024 · Shift 2 · Q43
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Motion question

2024 · Shift 2 · Q43

JEE AdvancedPhysicsMotionNumerical+4 / −1
A ball is thrown from the location (x0,y0)=(0,0)\left(x_0, y_0\right)=(0,0)(x0​,y0​)=(0,0) of a horizontal playground with an initial speed v0v_0v0​ at an angle θ0\theta_0θ0​ from the +x+x+x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x1,y1)=(L,0)\left(x_1, y_1\right)=(L, 0)(x1​,y1​)=(L,0). The stone is thrown at an angle (180−θ1)\left(180-\theta_1\right)(180−θ1​) from the +x+x+x-direction with a suitable initial speed. For a fixed v0v_0v0​, when (θ0,θ1)=(45∘,45∘)\left(\theta_0, \theta_1\right)=\left(45^{\circ}, 45^{\circ}\right)(θ0​,θ1​)=(45∘,45∘), the stone hits the ball after time T1T_1T1​, and when (θ0,θ1)=(60∘,30∘)\left(\theta_0, \theta_1\right)=\left(60^{\circ}, 30^{\circ}\right)(θ0​,θ1​)=(60∘,30∘), it hits the ball after time T2T_2T2​. In such a case, (T1/T2)2\left(T_1 / T_2\right)^2(T1​/T2​)2 is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Write the projectile equations

Let the ball be thrown from (0,0)(0,0)(0,0) with speed v0v_0v0​ at angle θ0\theta_0θ0​. Its coordinates at time ttt are

xb=v0cos⁡θ0 t,yb=v0sin⁡θ0 t−12gt2.x_b = v_0 \cos\theta_0\, t, \qquad y_b = v_0 \sin\theta_0\, t - \frac12 gt^2.xb​=v0​cosθ0​t,yb​=v0​sinθ0​t−21​gt2.

The stone is thrown from (L,0)(L,0)(L,0) at angle 180∘−θ1180^\circ-\theta_1180∘−θ1​, so its horizontal and vertical velocity components are

(−ucos⁡θ1,  usin⁡θ1),(-u\cos\theta_1,\; u\sin\theta_1),(−ucosθ1​,usinθ1​),

where uuu is its suitable initial speed. Thus,

xs=L−ucos⁡θ1 t,ys=usin⁡θ1 t−12gt2.x_s = L - u\cos\theta_1\, t, \qquad y_s = u\sin\theta_1\, t - \frac12 gt^2.xs​=L−ucosθ1​t,ys​=usinθ1​t−21​gt2.

For collision at time ttt, we need xb=xs,yb=ys.x_b=x_s, \qquad y_b=y_s.xb​=xs​,yb​=ys​.

  1. Use equality of vertical coordinates

From yb=ysy_b=y_syb​=ys​,

v0sin⁡θ0 t−12gt2=usin⁡θ1 t−12gt2.v_0\sin\theta_0\, t - \frac12 gt^2 = u\sin\theta_1\, t - \frac12 gt^2.v0​sinθ0​t−21​gt2=usinθ1​t−21​gt2.

For t>0t>0t>0, this gives

usin⁡θ1=v0sin⁡θ0.u\sin\theta_1 = v_0\sin\theta_0.usinθ1​=v0​sinθ0​.

So,

u=v0sin⁡θ0sin⁡θ1.u = \frac{v_0\sin\theta_0}{\sin\theta_1}.u=sinθ1​v0​sinθ0​​.
  1. Use equality of horizontal coordinates

From xb=xsx_b=x_sxb​=xs​,

v0cos⁡θ0 t=L−ucos⁡θ1 t.v_0\cos\theta_0\, t = L - u\cos\theta_1\, t.v0​cosθ0​t=L−ucosθ1​t.

So,

L=t(v0cos⁡θ0+ucos⁡θ1).L = t\left(v_0\cos\theta_0 + u\cos\theta_1\right).L=t(v0​cosθ0​+ucosθ1​).

Substitute u=v0sin⁡θ0sin⁡θ1u = \frac{v_0\sin\theta_0}{\sin\theta_1}u=sinθ1​v0​sinθ0​​ into this:

L=t(v0cos⁡θ0+v0sin⁡θ0sin⁡θ1cos⁡θ1).L = t\left(v_0\cos\theta_0 + \frac{v_0\sin\theta_0}{\sin\theta_1}\cos\theta_1\right).L=t(v0​cosθ0​+sinθ1​v0​sinθ0​​cosθ1​).

Hence,

t=Lv0(cos⁡θ0+sin⁡θ0cot⁡θ1).t = \frac{L}{v_0\left(\cos\theta_0 + \sin\theta_0\cot\theta_1\right)}.t=v0​(cosθ0​+sinθ0​cotθ1​)L​.

This gives the collision time for given (θ0,θ1)(\theta_0,\theta_1)(θ0​,θ1​).

  1. Case 1: (θ0,θ1)=(45∘,45∘)(\theta_0,\theta_1)=(45^\circ,45^\circ)(θ0​,θ1​)=(45∘,45∘)

Then

cos⁡45∘=sin⁡45∘=12,cot⁡45∘=1.\cos45^\circ = \sin45^\circ = \frac{1}{\sqrt2}, \qquad \cot45^\circ=1.cos45∘=sin45∘=2​1​,cot45∘=1.

So,

T1=Lv0(12+12)=L2 v0.T_1 = \frac{L}{v_0\left(\frac1{\sqrt2}+\frac1{\sqrt2}\right)} = \frac{L}{\sqrt2\,v_0}.T1​=v0​(2​1​+2​1​)L​=2​v0​L​.
  1. Case 2: (θ0,θ1)=(60∘,30∘)(\theta_0,\theta_1)=(60^\circ,30^\circ)(θ0​,θ1​)=(60∘,30∘)

Now,

cos⁡60∘=12,sin⁡60∘=32,cot⁡30∘=3.\cos60^\circ = \frac12, \qquad \sin60^\circ = \frac{\sqrt3}{2}, \qquad \cot30^\circ = \sqrt3.cos60∘=21​,sin60∘=23​​,cot30∘=3​.

Thus,

cos⁡60∘+sin⁡60∘cot⁡30∘=12+32⋅3=12+32=2.\cos60^\circ + \sin60^\circ\cot30^\circ = \frac12 + \frac{\sqrt3}{2}\cdot \sqrt3 = \frac12 + \frac32 = 2.cos60∘+sin60∘cot30∘=21​+23​​⋅3​=21​+23​=2.

Hence,

T2=L2v0.T_2 = \frac{L}{2v_0}.T2​=2v0​L​.
  1. Compute the required ratio
T1T2=L/(2v0)L/(2v0)=22=2.\frac{T_1}{T_2} = \frac{L/(\sqrt2 v_0)}{L/(2v_0)} = \frac{2}{\sqrt2} = \sqrt2.T2​T1​​=L/(2v0​)L/(2​v0​)​=2​2​=2​.

Therefore,

(T1T2)2=2.\left(\frac{T_1}{T_2}\right)^2 = 2.(T2​T1​​)2=2.
  1. Comparison with stored answer

Derived answer: 222. Stored correct answer: 222. They agree.

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