SATHEE: Kinematics 5 Question 16

Kinematics 5 Question 16

15. A cart is moving along $x$ -direction with a velocity of $4 m / s$ . A person on the cart throws a stone with a velocity of $6 m / s$ relative to himself. In the frame of reference of the cart, the stone is thrown in $y - z$ plane making an angle of $30^{\circ}$ with vertical $z$ -axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length $L$ .

A completely inelastic collision occurs, in which the stone gets embedded in the object. Determine $(g = 9.8 m / s^{2})$

(1997, 5M)

(a) the speed of the combined mass immediately after the collision with respect to an observer on the ground.

(b) the length $L$ of the string such that tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.

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Answer:

Correct Answer: 15. (a) $2.5 m / s$

(b) $0.32 m$

Solution:

(a) Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along $x, y$ and $z$ -directions respectively.

$\begin{aligned} Given, & v_{cart} & = 4 \hat{i} m / s \\ ∴ v_{stone, cart} & = (6 \sin 30^{\circ}) \hat{j} + (6 \cos 30^{\circ}) \hat{k} \\ = (3 \hat{j} + 3 \sqrt{3} \hat{k}) m / s \\ ∴ & v_{stone} & = v_{stone, cart} + v_{cart} \\ = (4 \hat{i} + 3 \hat{j} + 3 \sqrt{3} \hat{k}) m / s \end{aligned}$

This is the absolute velocity of stone (with respect to ground). At highest point of its trajectory, the vertical component $(z)$ of its velocity will become zero, whereas the $x$ and $y$ -components will remain unchanged. Therefore, velocity of stone at highest point will be,

$v = (4 \hat{i} + 3 \hat{j}) m / s$

or speed at highest point,

$v = | v | = (\sqrt{(4)^{2} + (3)^{2}} m / s = 5 m / s$

Now, applying law of conservation of linear momentum, let $v_{0}$ be the velocity of combined mass after collision.

$\begin{array}{ll} Then, & m v = (2 m) v_{0} \\ ∴ & v_{0} = \frac{v}{2} = \frac{5}{2} m / s = 2.5 m / s \end{array}$

$∴$ Speed of combined mass just after collision is $2.5 m / s$ .

(b) Tension in the string becomes zero at horizontal position. It implies that velocity of combined mass also becomes zero in horizontal position. Applying conservation of energy, we have