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A metal ball of mass 2 kg moving with speed of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after collision, both the balls move as a single mass, then the loss in K.E. due to collision is
To calculate the loss in kinetic energy due to the collision, we first calculate the initial and final kinetic energies of the system. Step 1: Convert initial speed to m/s Initial speed of the moving ball = 36 km/h Speed in m/s = (36 × 1000) / (60 × 60) = 10 m/s Step 2: Calculate the initial kineticRead more
To calculate the loss in kinetic energy due to the collision, we first calculate the initial and final kinetic energies of the system.
Step 1: Convert initial speed to m/s
Initial speed of the moving ball = 36 km/h
Speed in m/s = (36 × 1000) / (60 × 60) = 10 m/s
Step 2: Calculate the initial kinetic energy (K.E.) of the system
Only the first ball is moving initially, so:
Initial K.E. = (1/2) m₁ v₁²
Where:
– m₁ = 2 kg (the moving ball mass)
– v₁ = 10 m/s (the velocity of the moving ball)
Initial K.E. = (1/2) × 2 × (10)²
Initial K.E. = 100 J
.
Step 3: Calculating the resulting velocity of the merged mass
Immediately after impact, the balls have merged as a single object mass
– Total mass = m₁ + m₂ = 2 kg + 3 kg = 5 kg
Applying the law of conservation of momentum:
Initial momentum = Final momentum
m₁ v₁ + m₂ v₂ = (m₁ + m₂) v
where:
– v₂ = 0 (a stationary ball)
Substitute:
(2 x 10) + (3 x 0) = 5v
20 = 5v
v = 4 m/s
Step 4: Find the final kinetic energy of the system (K.E.)
Final K.E. = (1/2) (m₁ + m₂) v²
Final K.E. = (1/2) × 5 × (4)²
Final K.E. = (1/2) × 5 × 16
Final K.E. = 40 J
Step 5: Calculate the loss in kinetic energy
Loss in K.E. = Initial K.E. − Final K.E.
Loss in K.E. = 100 J − 40 J
Loss in K.E. = 60 J
Final Answer:
Loss in kinetic energy due to collision is 60 J.
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How does the escape velocity on the Moon compare to that on Earth?
The escape velocity of the Moon is smaller due to its smaller mass and less radius than Earth. This is a question of Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/ncert-solutions/Read more
The escape velocity of the Moon is smaller due to its smaller mass and less radius than Earth. This is a question of Chapter 7 physics Class 11th NCERT. From the Chapter 7. Gravitation. Answer according to your understanding.
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250 N force is required to raise 75 kg mass from a pulley. If rope is pulled 12 m, then the load is lifted to 3 m, the efficiency of pulley system will be
We use the formula for efficiency of a pulley system: Efficiency = (Useful Work Output / Total Work Input) × 100 Step 1: Calculate Useful Work Output Useful work output is the work done in lifting the load. It is given by: Useful Work Output = Force × Distance lifted Where, - Force or weight of theRead more
We use the formula for efficiency of a pulley system:
Efficiency = (Useful Work Output / Total Work Input) × 100
Step 1: Calculate Useful Work Output
Useful work output is the work done in lifting the load. It is given by:
Useful Work Output = Force × Distance lifted
Where,
– Force or weight of the load = mass × gravitational acceleration
– Weight of the load = 75 kg × 9.8 m/s² = 735 N
– Distance lifted = 3 m
Therefore, the useful work output is:
Work Output = 735 N × 3 m = 2205 J
Step 2: Calculate Total Work Input
The total work input is the work done in pulling the rope. This is calculated as:
Total Work Input = Force applied × Distance pulled
Where:
– Force applied = 250 N
– Distance pulled = 12 m
Therefore, the total work input is:
Work Input = 250 N × 12 m = 3000 J
Step 3: Calculate Efficiency
Now, we can find the efficiency:
Efficiency = (Work Output / Work Input) × 100
Efficiency = (2205 J / 3000 J) × 100 ≈ 73.5%
The closest option is 75%.
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How much water a pump of 2 kW can raise in one minute to a height of 10 m? (take g = 10m /s²)
To determine how much water a 2 kW pump can raise in one minute to a height of 10 m, we can use the formula for power: P = W / t Step 1: Calculate Work Done The work done to raise water to a height h is given by: W = mgh Where: - m is the mass of the water in kilograms (kg) - g is the acceleration dRead more
To determine how much water a 2 kW pump can raise in one minute to a height of 10 m, we can use the formula for power:
P = W / t
Step 1: Calculate Work Done
The work done to raise water to a height h is given by:
W = mgh
Where:
– m is the mass of the water in kilograms (kg)
– g is the acceleration due to gravity (10 m/s²)
– h is the height in meters (10 m)
Step 2: Convert Power to Work Done in One Minute
Given:
– Power, P = 2 kW = 2000 W
– Time, t = 1 minute = 60 s
Now calculate the work done:
W = P × t
W = 2000 W × 60 s
W = 120000 J
Step 3: Calculate the Mass of Water
Now we can use the work done to calculate the mass of water:
W = mgh => m = W / (gh)
Substitute known values:
m = 120000 J / (10 m/s² × 10 m)
m = 120000 / 100
m = 1200 kg
Step 4: Convert Mass to Volume
Knowing that the density of water is around 1000 kg/m³, the volume V of water lifted will be:
V = m / density
V = 1200 kg / (1000 kg/m³)
V = 1.2 m³
Converting the above value to cubic meters into liters:
Since,
1 m³ = 1000 liters,
V = 1.2 m³ × 1000 liters/m³
V = 1200 liters
Final Answer:
In one minute, the pump is able to pump 1200 liters of water up a height of 10 m.
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The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is
To determine the potential energy in a stretched spring, we have the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension (or compression) of the spring from its equilibrium position Given: - When the spring is stretcRead more
To determine the potential energy in a stretched spring, we have the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension (or compression) of the spring from its equilibrium position
Given:
– When the spring is stretched by x₁ = 2 cm:
U = (1/2) k (2 cm)² = (1/2) k (0.02 m)²
– When the spring is stretched by x₂ = 8 cm:
U’ = (1/2) k (8 cm)² = (1/2) k (0.08 m)²
Step 1: Calculate the ratio of potential energies
We are required to find the ratio of the potential energies when stretched by 8 cm compared to when stretched by 2 cm:
(U’) / (U) = [(1/2) k (0.08 m)²] / [(1/2) k (0.02 m)²] = [(0.08)²] / [(0.02)²]
Step 2: Simplify the ratio
Compute the squares:
(U’) / (U) = [(0.08)²] / [(0.02)²] = [0.0064] / [0.0004] = 16
Conclusion
The potential energy in the spring stretched by 8 cm is given by:
U’ = 16 U
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