SATHEE: Kinematics 5 Question 18

Kinematics 5 Question 18

17. Two towers $A B$ and $C D$ are situated a distance $d$ apart as shown in figure. $A B$ is $20 m$ high and $C D$ is $30 m$ high from the ground. An object of mass $m$ is thrown from the top of $A B$ horizontally with a velocity of $10 m / s$ towards $C D$ .

Simultaneously, another object of mass $2 m$ is thrown from the top of $C D$ at an angle of $60^{\circ}$ to the horizontal towards $A B$ with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick to each other.

$(1994, 6$ M)

(a) Calculate the distance $d$ between the towers.

(b) Find the position where the objects hit the ground.

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Answer:

Correct Answer: 17. (a) Approximately $17.32 m$

(b) $11.55 m$ from $B$ 18. $\frac{1}{2}$

Solution:

(a) Acceleration of $A$ and $C$ both is $9.8 m / s^{2}$ downwards. Therefore, relative acceleration between them is zero i.e. the relative motion between them will be uniform.

Now, assuming $A$ to be at rest, the condition of collision will be that $v_{C A} = v_{C} - v_{A} =$ relative velocity of $C$ w.r.t. $A$ should be along $C A$ .

$\begin{array}{ll} ∴ & \frac{{(v_{C A})}_{y}}{{(v_{C A})}_{x}} = \frac{C E}{A E} = \frac{10}{d} \\ or \frac{v_{C y} - v_{A y}}{v_{C x} - v_{A x}} = \frac{10}{d} or \frac{5 \sqrt{3} - 0}{5 - (- 10)} = \frac{10}{d} \\ ∴ & d = 10 \sqrt{3} m \approx 17.32 m \end{array}$

(b) Time of collision, $t = \frac{A C}{∣ v_{C A}}$

$\begin{aligned} | v_{C A} | & = \sqrt{(v_{C A x}^{2}) + {(v_{C A y})}^{2}} \\ = \sqrt{{5 - (- 10)}^{2} + {5 \sqrt{3} - 0}^{2}} = 10 \sqrt{3} m / s \\ C A & = \sqrt{(10)^{2} + (10 \sqrt{3})^{2}} = 20 m \\ ∴ t & = \frac{20}{10 \sqrt{3}} = \frac{2}{\sqrt{3}} s \end{aligned}$

Horizontal (or $x$ ) component of momentum of $A$ , i.e. $p_{A x} = m v_{A x} = - 10 m$ .

Similarly, $x$ component of momentum of $C$ , i.e.

Since,

$p_{C x} = (2 m) v_{C x} = (2 m) (5) = + 10 m$

i.e. $x$ -component of momentum of combined mass after collision will also be zero, i.e. the combined mass will have momentum or velocity in vertical or $y$ -direction only.

Hence, the combined mass will fall at point $F$ just below the point of collision $P$ .

Here

$\begin{array}{ll} Here & d_{1} = | (v_{A x}) | t = (10) \frac{2}{\sqrt{3}} = 11.55 m \\ ∴ & d_{2} = (d - d_{1}) = (17.32 - 11.55) = 5.77 m \end{array}$