We need to find the distance Truck A travels before stopping. We use the kinematic equation $v^2 = u^2 + 2as$, where:
Plugging in the values:
$0^2 = (16 \ m/s)^2 + 2(-2 \ m/s^2)s_A$
$0 = 256 \ m^2/s^2 - (4 \ m/s^2)s_A$
$s_A = \frac{256 \ m^2/s^2}{4 \ m/s^2} = 64 \ m$
Truck A travels $64 \ m$ before stopping.
Similarly, we find the distance Truck B travels before stopping using the same kinematic equation $v^2 = u^2 + 2as$. For Truck B:
Substituting the values:
$0^2 = (20 \ m/s)^2 + 2(-4 \ m/s^2)s_B$
$0 = 400 \ m^2/s^2 - (8 \ m/s^2)s_B$
$s_B = \frac{400 \ m^2/s^2}{8 \ m/s^2} = 50 \ m$
Truck B travels $50 \ m$ before stopping.
The initial distance between the trucks was $200 \ m$. When they stop, the total distance covered by both trucks is the sum of their stopping distances.
Total distance covered $= s_A + s_B = 64 \ m + 50 \ m = 114 \ m$.
The final distance between them is the initial distance minus the total distance they covered while braking.
Final distance $= \text{Initial Distance} - (s_A + s_B)$
Final distance $= 200 \ m - 114 \ m = 86 \ m$.
Therefore, the distance between the trucks when they stop is $86 \ m$.
A uniform bar of length 12 cm and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of $v$ and $\omega$ is :
Match the LIST-I with LIST-II
| List-I: | List-II: |
| A. Magnetic induction | I. $[M L T^{-2} A^{-2}]$ |
| B. Magnetic flux | II. $[M L^2 T^{-2} A^{-2}]$ |
| C. Magnetic permeability | III. $[M L^0 T^{-2} A^{-1}]$ |
| D. Self inductance | IV. $[M L^2 T^{-2} A^{-1}]$ |
Choose the correct answer from the options given below:
In the given figure the blocks $A$, $B$ and $C$ weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $\vec{F}$ required to slide the block $C$ with constant speed is ______ N. (Use $g = 10 \text{ m/s}^2$)

In case of vertical circular motion of a particle by a thread of length $r$ if the tension in the thread is zero at an angle $30^\circ$ shown in figure, the velocity at the bottom point ($A$) of the circular path is
($g = \text{gravitational acceleration}$)

A uniform bar of length 12 cm and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of $v$ and $\omega$ is :
Match the LIST-I with LIST-II
| List-I: | List-II: |
| A. Magnetic induction | I. $[M L T^{-2} A^{-2}]$ |
| B. Magnetic flux | II. $[M L^2 T^{-2} A^{-2}]$ |
| C. Magnetic permeability | III. $[M L^0 T^{-2} A^{-1}]$ |
| D. Self inductance | IV. $[M L^2 T^{-2} A^{-1}]$ |
Choose the correct answer from the options given below: