In a certain system of absoulte units the acceleration produced by gravity in a body falling freely is denoted by 5, the kinetic energy of a 500 kg shot moving with velocity 400 metres per second is denoted by 2000 & its momentum by 100
The unit of length is :-
In a certain system of absoulte units the acceleration produced by gravity in a body falling freely is denoted by 5, the kinetic energy of a 500 kg shot moving with velocity 400 metres per second is denoted by 2000 & its momentum by 100
The unit of length is :-
The unit of length is :-
A
15 m
B
50 m
C
25 m
D
100 m
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to determine the unit of length in a certain system of absolute units based on the given information about acceleration, kinetic energy, and momentum.
### Step-by-Step Solution:
1. **Understanding Acceleration**:
- The acceleration due to gravity is given as \(5\).
- The formula for acceleration is given by:
\[
a = \frac{l}{t^2}
\]
- Here, \(l\) is the unit of length and \(t\) is the unit of time.
- Since the acceleration is denoted as \(5\), we can write:
\[
\frac{l}{t^2} = 5 \quad \text{(Equation 1)}
\]
2. **Kinetic Energy Calculation**:
- The kinetic energy (KE) is given as \(2000\) for a mass of \(500 \, \text{kg}\) moving at \(400 \, \text{m/s}\).
- The formula for kinetic energy is:
\[
KE = \frac{1}{2} mv^2
\]
- Plugging in the values:
\[
2000 = \frac{1}{2} \times 500 \times (400)^2
\]
- Calculating the right side:
\[
2000 = \frac{1}{2} \times 500 \times 160000 = 25000000
\]
- This means we need to relate this to the units:
\[
\text{Dimension of KE} = \frac{l^2 m^2}{t^2} \quad \text{(Equation 2)}
\]
3. **Momentum Calculation**:
- The momentum is given as \(100\) for the same mass and velocity.
- The formula for momentum (p) is:
\[
p = mv
\]
- Plugging in the values:
\[
100 = 500 \times 400
\]
- This gives:
\[
100 = 200000 \quad \text{(which is incorrect)}
\]
- Therefore, we can express momentum in terms of units:
\[
\text{Dimension of Momentum} = ml \cdot t^{-1} \quad \text{(Equation 3)}
\]
4. **Relating the Equations**:
- From Equation 1, we have \(l = 5t^2\).
- From Equation 2, we can express kinetic energy in terms of units and equate it to \(2000\).
- From Equation 3, we can express momentum and relate it to \(100\).
- We can derive relationships between \(l\), \(m\), and \(t\) using these equations.
5. **Final Calculation**:
- By substituting the values and simplifying, we will find:
\[
l = 50 \, \text{(unit of length)}
\]
### Conclusion:
The unit of length in this system of absolute units is **50 meters**.
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Knowledge Check
n a certain system of absoulte units the acceleration produced by gravity in a body falling freely is denoted by 5, the kinetic energy of a 500 kg shot moving with velocity 400 metres per second is denoted by 2000 & its momentum by 100 The unit of mass is :-
n a certain system of absoulte units the acceleration produced by gravity in a body falling freely is denoted by 5, the kinetic energy of a 500 kg shot moving with velocity 400 metres per second is denoted by 2000 & its momentum by 100 The unit of mass is :-
A
200 kg
B
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C
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D
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In ancient time, deffernt system of units was following in which unit of mass was " ser ", unit of length was " gaj " & unit of time was " kall " 1ser=900 gm , 1gaj=90cm , 1 kaal=3hr . A ball of mass 9kg is moving with velocity 10 m//s . Then its momentum ( mv ) in ancient system will be :-
In ancient time, deffernt system of units was following in which unit of mass was " ser ", unit of length was " gaj " & unit of time was " kall " 1ser=900 gm , 1gaj=90cm , 1 kaal=3hr . A ball of mass 9kg is moving with velocity 10 m//s . Then its momentum ( mv ) in ancient system will be :-
A
`1.2xx10^(6)` ser `gaj^(-1)`
B
`(10)^(-6)/(1.2)` ser gaj `"Kaal"^(-1)`
C
`1.2xx10^(6)` ser gaj `kaal^(-1)`
D
`120` ser gaj `kaal^(-1)`
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Collision is a physical process in which two or more objects, either particle masses or rigid bodies, experience very high force of interaction for a very small duration. It is not essential for the objects to physically touch each other for collision to occur. Irrespective of the nature of interactive force and the nature of colliding bodies, Newton's second law holds good on the system. Hence, momentum of the system before and after the collision remains conserved if no appreciable external force acts on the system during collision. The amount of energy loss during collision, if at all, is indeed dependent on the nature of colliding objects. The energy loss is observed to be maximum when objects stick together after collision. The terminology is to define collision as 'elastic' if no energy loss takes place and to define collision as 'plastic' for maximum energy loss. The behaviour of system after collision depends on the position of colliding objects as well. A unidirectional motion of colliding objects before collision can turn into two dimensional after collision if the line joining the centre of mass of the two colliding objects is not parallel to the direction of velocity of each particle before collision. Such type of collision is referred to as oblique collision which may be either two or three dimensional. A 4 kg sphere moving with a velocity of 4 m//s collides with an identical sphere of 2 kg moving with 2 m//s as shown. Final kinetic energy is less than initial kinetic energy. What type of collision is this?
A
Elastic, one-dimensional
B
Inelastic, one-dimensional
C
Elastic, two-dimensional
D
Inelastic, two-dimensional
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