JEE Challenger
More from Motion in a Plane
PhyMedAdv 2024 • Paper 2

Ratio of Collision Times for Two Projectiles

A ball is thrown from the location (x0,y0)=(0,0)(x_0, y_0) = (0,0) of a horizontal playground with an initial speed v0v_0 at an angle θ0\theta_0 from the +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)(x_1, y_1) = (L, 0). The stone is thrown at an angle (180−θ1)(180 - \theta_1) from the +x+x-direction with a suitable initial speed. For a fixed v0v_0, when (θ0,θ1)=(45∘,45∘)(\theta_0, \theta_1) = (45^\circ, 45^\circ), the stone hits the ball after time T1T_1, and when (θ0,θ1)=(60∘,30∘)(\theta_0, \theta_1) = (60^\circ, 30^\circ), it hits the ball after time T2T_2. In such a case, (T1/T2)2(T_1/T_2)^2 is _____.

Official Numerical Answer2

Topics & Concepts

Motion in a PlaneProjectile MotionRelative Motion

Step-by-Step Solution

To find the ratio of the collision times (T1/T2)2(T_1/T_2)^2, we analyze the relative motion and position equations of the ball and the stone.

1. Equations of Motion for the Ball and the Stone

Let the ball be thrown from (x0,y0)=(0,0)(x_0, y_0) = (0,0) with initial velocity v0v_0 at an angle θ0\theta_0 with the +x+x-axis. Its position as a function of time tt is given by: xB(t)=v0cos⁡θ0 tx_B(t) = v_0 \cos\theta_0 \, t yB(t)=v0sin⁡θ0 t−12gt2y_B(t) = v_0 \sin\theta_0 \, t - \frac{1}{2}gt^2

The stone is thrown from (x1,y1)=(L,0)(x_1, y_1) = (L,0) at the same time with initial speed v1v_1 at an angle (180∘−θ1)(180^\circ - \theta_1) from the +x+x-axis (i.e., at an angle θ1\theta_1 above the −x-x-axis). Its position as a function of time tt is: xS(t)=L−v1cos⁡θ1 tx_S(t) = L - v_1 \cos\theta_1 \, t yS(t)=v1sin⁡θ1 t−12gt2y_S(t) = v_1 \sin\theta_1 \, t - \frac{1}{2}gt^2

2. Collision Condition

For the stone to hit the ball at time TT, their coordinates must be identical at t=Tt = T:

  1. Equating vertical coordinates (yB(T)=yS(T)y_B(T) = y_S(T)): v0sin⁡θ0 T−12gT2=v1sin⁡θ1 T−12gT2v_0 \sin\theta_0 \, T - \frac{1}{2}gT^2 = v_1 \sin\theta_1 \, T - \frac{1}{2}gT^2 v0sin⁡θ0=v1sin⁡θ1  ⟹  v1=v0sin⁡θ0sin⁡θ1v_0 \sin\theta_0 = v_1 \sin\theta_1 \implies v_1 = v_0 \frac{\sin\theta_0}{\sin\theta_1}

  2. Equating horizontal coordinates (xB(T)=xS(T)x_B(T) = x_S(T)): v0cos⁡θ0 T=L−v1cos⁡θ1 Tv_0 \cos\theta_0 \, T = L - v_1 \cos\theta_1 \, T (v0cos⁡θ0+v1cos⁡θ1)T=L(v_0 \cos\theta_0 + v_1 \cos\theta_1) T = L

Substitute v1=v0sin⁡θ0sin⁡θ1v_1 = v_0 \frac{\sin\theta_0}{\sin\theta_1} into the equation above: v0(cos⁡θ0+sin⁡θ0cos⁡θ1sin⁡θ1)T=Lv_0 \left( \cos\theta_0 + \frac{\sin\theta_0 \cos\theta_1}{\sin\theta_1} \right) T = L v0(sin⁡θ1cos⁡θ0+sin⁡θ0cos⁡θ1sin⁡θ1)T=Lv_0 \left( \frac{\sin\theta_1 \cos\theta_0 + \sin\theta_0 \cos\theta_1}{\sin\theta_1} \right) T = L

Using the trigonometric identity sin⁡(θ0+θ1)=sin⁡θ0cos⁡θ1+cos⁡θ0sin⁡θ1\sin(\theta_0 + \theta_1) = \sin\theta_0 \cos\theta_1 + \cos\theta_0 \sin\theta_1: v0sin⁡(θ0+θ1)sin⁡θ1T=Lv_0 \frac{\sin(\theta_0 + \theta_1)}{\sin\theta_1} T = L T=Lv0sin⁡θ1sin⁡(θ0+θ1)T = \frac{L}{v_0} \frac{\sin\theta_1}{\sin(\theta_0 + \theta_1)}

3. Calculation for Both Cases

  • Case 1: (θ0,θ1)=(45∘,45∘)(\theta_0, \theta_1) = (45^\circ, 45^\circ) T1=Lv0sin⁡45∘sin⁡(45∘+45∘)=Lv01/2sin⁡90∘=L2v0T_1 = \frac{L}{v_0} \frac{\sin 45^\circ}{\sin(45^\circ + 45^\circ)} = \frac{L}{v_0} \frac{1/\sqrt{2}}{\sin 90^\circ} = \frac{L}{\sqrt{2} v_0}

  • Case 2: (θ0,θ1)=(60∘,30∘)(\theta_0, \theta_1) = (60^\circ, 30^\circ) T2=Lv0sin⁡30∘sin⁡(60∘+30∘)=Lv01/2sin⁡90∘=L2v0T_2 = \frac{L}{v_0} \frac{\sin 30^\circ}{\sin(60^\circ + 30^\circ)} = \frac{L}{v_0} \frac{1/2}{\sin 90^\circ} = \frac{L}{2 v_0}

4. Ratio of Collision Times

T1T2=L2v0L2v0=22=2\frac{T_1}{T_2} = \frac{\frac{L}{\sqrt{2}v_0}}{\frac{L}{2v_0}} = \frac{2}{\sqrt{2}} = \sqrt{2}

Squaring both sides: (T1T2)2=(2)2=2\left( \frac{T_1}{T_2} \right)^2 = (\sqrt{2})^2 = 2

PreviousNext
Ratio of Collision Times for Two Projectiles | Physics PYQ Solution - JEE Challenger