Solution for Question 44
Vector Analysis
The velocity of the disc relative to the ground is the vector sum of the belt’s velocity and the disc’s relative velocity:
$$ \vec{v}_{ground} = \vec{v}_{belt} + \vec{v}_{rel}(t) $$Here, $\vec{v}_{belt} = 3\hat{i}$. The relative velocity $\vec{v}_{rel}$ starts at $5$ m/s pointing backwards and upwards (direction vectors $-3\hat{i} + 4\hat{j}$) and decreases linearly to zero due to friction.
We can parametrize the relative velocity using a parameter $k$ which goes from 1 to 0 (where $k$ represents the fraction of initial speed remaining):
$$ \vec{v}_{rel} = k(-3\hat{i} + 4\hat{j}) $$ $$ \vec{v}_{ground} = 3\hat{i} + k(-3\hat{i} + 4\hat{j}) = (3 – 3k)\hat{i} + 4k\hat{j} $$Minimization
We need to minimize the speed $|\vec{v}_{ground}|$. Let’s minimize the square of the speed $V^2$:
$$ V^2 = (3 – 3k)^2 + (4k)^2 $$ $$ V^2 = 9(1-k)^2 + 16k^2 $$Differentiating with respect to $k$ and setting to zero:
$$ \frac{d(V^2)}{dk} = 18(1-k)(-1) + 32k = 0 $$ $$ -18 + 18k + 32k = 0 $$ $$ 50k = 18 \implies k = \frac{18}{50} = 0.36 $$Now substitute $k = 0.36$ back into the velocity components:
$$ v_x = 3(1 – 0.36) = 3(0.64) = 1.92 $$ $$ v_y = 4(0.36) = 1.44 $$ $$ V_{min} = \sqrt{(1.92)^2 + (1.44)^2} $$Alternatively, substitute $k$ directly into the squared equation:
$$ V^2 = 9(0.64)^2 + 16(0.36)^2 = 3.6864 + 2.0736 = 5.76 $$ $$ V_{min} = \sqrt{5.76} = 2.4 \text{ m/s} $$Answer: (c) 2.4 m/s