1. To get a resultant displacement of $10\text{ m}$, two displacement vectors of magnitude $6\text{ m}$ and $8\text{ m}$ should be combined:
Verification Matrix:
Using the vector addition formula, $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$. Here, $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ m}$. This holds true only when $\cos\theta = 0$, which means $\theta = 90^\circ$ (perpendicular to each other).
2. The velocity of a particle at an instant is $10\text{ m/s}$ and after $5\text{ s}$ the velocity of the particle is $20\text{ m/s}$. The velocity $3\text{ s}$ before in $\text{m/s}$ was:
Verification Matrix:
First, find uniform acceleration: $a = \frac{v_f - v_i}{t} = \frac{20 - 10}{5} = 2\text{ m/s}^2$. To find the velocity $3\text{ s}$ before the $10\text{ m/s}$ mark, look backward in time ($t = -3\text{ s}$): $v = v_i + at = 10 + (2)(-3) = 10 - 6 = 4\text{ m/s}$. Correction note: Based on standard linear translation mechanics, the calculation gives $4\text{ m/s}$ (Option b). Let's trace if a typo exists in the key option configuration, but mathematically it resolves to 4.
3. A ball is thrown upwards with a velocity of $100\text{ m/s}$. It will reach the ground after:
Verification Matrix:
Total time of flight for a vertical projectile is given by $T = \frac{2v_i}{g}$. Assuming $g \approx 10\text{ m/s}^2$, we calculate $T = \frac{2 \times 100}{10} = 20\text{ seconds}$. (It takes $10\text{ s}$ to go up and $10\text{ s}$ to return).
4. Two projectiles are fired from the same point with the same speed at angles of projection $60^\circ$ and $30^\circ$ respectively. Which one of the following is true?
Verification Matrix:
Projectiles fired with identical initial velocity at complementary angles (angles that sum to $90^\circ$, such as $30^\circ$ and $60^\circ$) map to the exact same horizontal range because $\sin(2 \times 30^\circ) = \sin(2 \times 60^\circ) = \sin(60^\circ)$.
5. The ratio of numerical values of average velocity and average speed of a body is always:
Verification Matrix:
Displacement is always less than or equal to distance ($\text{Displacement} \le \text{Distance}$). Since $\text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}}$ and $\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}$, the ratio $\frac{\text{Average Velocity}}{\text{Average Speed}}$ must be equal to 1 (unity) for a straight line paths or less than 1 if turning occurs.
6. If the average velocities of a body become equal to the instantaneous velocity, the body is said to be moving with:
Verification Matrix:
When a body moves with uniform velocity, its speed and direction do not change over time. Thus, its velocity at any specific split second (instantaneous) matches the calculation taken over any extended period (average).
7. At the top of a trajectory of a projectile, the acceleration is:
Verification Matrix:
Neglecting air resistance, a projectile is uniquely acted upon by gravity throughout its entire journey. Even at the peak of its trajectory where vertical velocity drops to zero, downward acceleration due to gravity remains constant at $g$.
8. At what angle the range of projectile becomes equal to the height of projectile?
Verification Matrix:
Setting Range equal to Height ($R = H$): $\frac{v_i^2\sin(2\theta)}{g} = \frac{v_i^2\sin^2\theta}{2g} \implies 2\sin\theta\cos\theta = \frac{\sin^2\theta}{2}$. Simplifying yields $\tan\theta = 4$. Taking the inverse, $\theta = \tan^{-1}(4) \approx 76^\circ$.
9. The angle at which dot product becomes equal to the cross product magnitude is:
Verification Matrix:
The dot product is given by $AB\cos\theta$ and the magnitude of the cross product is $AB\sin\theta$. Equating them: $AB\cos\theta = AB\sin\theta \implies \tan\theta = 1$. This yields $\theta = 45^\circ$.
10. If the dot product of two non-zero vectors vanishes; the vectors will be:
Verification Matrix:
The dot product is defined as $\vec{A} \cdot \vec{B} = AB\cos\theta$. For two non-zero vectors ($A \neq 0, B \neq 0$), the product can only vanish (equal 0) if $\cos\theta = 0$, which occurs when the vectors are perpendicular ($\theta = 90^\circ$).
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