The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance betRead more
The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
For two objects of masses m1 and m2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:
𝐹=𝐺𝑚¹𝑚²/𝑟²
Where, G is the universal gravitation constant and its value is 6.67×10⁻¹¹ 𝑁𝑚 ²𝑘𝑔⁻ ².
Mass of the dumbbell, m = 10 kg Distance covered by the dumbbell, s = 80 cm = 0.8 m Acceleration in the downward direction, a = 10 m/s² Initial velocity of the dumbbell, u = 0 Final velocity of the dumbbell (when it was about to hit the floor) = v According to the third equation of motion: v²= u² +Read more
Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s²
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v²= u² + 2as
v²= 0 + 2 (10) 0.8
v= 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv
= 10 × 4 kg m s⁻¹
= 40 kg m s⁻¹
Initial velocity of the object, u = 5 m/s Final velocity of the object, v = 8 m/s Mass of the object, m = 100 kg Time take by the object to accelerate, t = 6 s Initial momentum = mu = 100 × 5 = 500 kg ms ⁻¹ Final momentum = mv= 100 × 8 = 800 kg ms⁻¹ Force exerted on the object, 𝐹 = 𝑚𝑣−𝑚𝑢/𝑡 = m(v-u)/Read more
Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg ms ⁻¹
Final momentum = mv= 100 × 8 = 800 kg ms⁻¹
Force exerted on the object, 𝐹 = 𝑚𝑣−𝑚𝑢/𝑡
= m(v-u)/t = 800-500/6 = 300/6 = 50N
Initial momentum of the object is 500 kg ms⁻¹
.
Final momentum of the object is 800 kg ms-1.
Force exerted on the object is 50 N.
Mass of the object, m1 = 1 kg Velocity of the object before collision, v₁ = 10 m/s Mass of the stationary wooden block, m₂ = 5 kg Velocity of the wooden block before collision, v₂= 0 m/s ∴ Total momentum before collision = m₁ v₁ + m₂ v₂ = 1 (10) + 5 (0) = 10 kg m s ⁻¹ It is given that after collisioRead more
Mass of the object, m1 = 1 kg
Velocity of the object before collision, v₁ = 10 m/s
Mass of the stationary wooden block, m₂ = 5 kg
Velocity of the wooden block before collision, v₂= 0 m/s
∴ Total momentum before collision = m₁ v₁ + m₂ v₂
= 1 (10) + 5 (0) = 10 kg m s ⁻¹
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁ v₁ + m₂ v₂ = (m1+ m₂) v
1 (10) + 5 (0) = (1 + 5) v
V=10/6 = 5/3 m/s
The total momentum after collision = 10 and velocity of combined object is 5/3
m/s.
Now, it is given that the bullet is travelling with a velocity of 150 m/s. Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s Final velocity, v = 0 (since the bullet finally comes to rest) Time taken to come to rest, t= 0.03 s According to the first equation of motiRead more
Now, it is given that the bullet is travelling with a velocity of 150 m/s.
Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v = u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)
A = -150/0.03 = – 5000 m/s²
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v²= u² + 2as
0 = (150)² + 2 ( – 5000) s
S = – (150)²/-2(5000) = 22500/10000 = 2.25m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass x Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s²
F = ma = 0.01×5000 = 50 N
Mass of the hockey ball, m= 200 g = 0.2 kg Hockey ball travels with velocity, v₁= 10 m/s Initial momentum = mv₁ Hockey ball travels in the opposite direction with velocity, v₂= -5 m/s Final momentum = mv₂ Change in momentum = mv₁- mv₂= 0.2 [10 - (-5)] = 0.2 (15) = 3 kg m s⁻¹ Hence, the change in momRead more
Mass of the hockey ball, m= 200 g = 0.2 kg
Hockey ball travels with velocity, v₁= 10 m/s
Initial momentum = mv₁
Hockey ball travels in the opposite direction with velocity, v₂= -5 m/s
Final momentum = mv₂
Change in momentum = mv₁- mv₂= 0.2 [10 – (-5)] = 0.2 (15) = 3 kg m s⁻¹
Hence, the change in momentum of the hockey ball is 3 kg m s⁻¹.
The truck has a large mass. Therefore, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in onRead more
The truck has a large mass. Therefore, the static friction between the truck and
the road is also very high. To move the car, one has to apply a force more than
the static friction. Therefore, when someone pushes the truck and the truck does
not move, then it can be said that the applied force in one direction is cancelled
out by the frictional force of equal amount acting in the opposite direction.
Therefore, the student is right in justifying that the two opposite and equal cancel
each other.
Mass of one of the objects, m₁ = 1.5 kg Mass of the other object, m₂ = 1.5 kg Velocity of m1 before collision, v₁ = 2.5 m/s Velocity of m₂, moving in opposite direction before collision, v₂ = -2.5 m/s (Negative sign arises because mass m₂ is moving in an opposite direction) After collision, the twoRead more
Mass of one of the objects, m₁ = 1.5 kg
Mass of the other object, m₂ = 1.5 kg
Velocity of m1 before collision, v₁ = 2.5 m/s
Velocity of m₂, moving in opposite direction before collision, v₂ = -2.5 m/s
(Negative sign arises because mass m₂ is moving in an opposite direction)
After collision, the two objects stick together.
Total mass of the combined object = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁ + v₁ + m₂ v₂= (m₁ + m₂) v
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5) v
3.75 – 3.75 = 3 v
v = 0
Hence, the velocity of the combined object after collision is 0 m/s.
A force of 200 N is applied in the forward direction. Thus, from Newton's third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the fictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet. For more answerRead more
A force of 200 N is applied in the forward direction. Thus, from Newton’s third
law of motion, an equal amount of force will act in the opposite direction. This
opposite force is the fictional force exerted on the cabinet. Hence, a frictional
force of 200 N is exerted on the cabinet.
State the universal law of gravitation.
The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance betRead more
The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
For two objects of masses m1 and m2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:
𝐹=𝐺𝑚¹𝑚²/𝑟²
Where, G is the universal gravitation constant and its value is 6.67×10⁻¹¹ 𝑁𝑚 ²𝑘𝑔⁻ ².
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How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s–2.
Mass of the dumbbell, m = 10 kg Distance covered by the dumbbell, s = 80 cm = 0.8 m Acceleration in the downward direction, a = 10 m/s² Initial velocity of the dumbbell, u = 0 Final velocity of the dumbbell (when it was about to hit the floor) = v According to the third equation of motion: v²= u² +Read more
Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s²
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v²= u² + 2as
v²= 0 + 2 (10) 0.8
v= 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv
= 10 × 4 kg m s⁻¹
= 40 kg m s⁻¹
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An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Initial velocity of the object, u = 5 m/s Final velocity of the object, v = 8 m/s Mass of the object, m = 100 kg Time take by the object to accelerate, t = 6 s Initial momentum = mu = 100 × 5 = 500 kg ms ⁻¹ Final momentum = mv= 100 × 8 = 800 kg ms⁻¹ Force exerted on the object, 𝐹 = 𝑚𝑣−𝑚𝑢/𝑡 = m(v-u)/Read more
Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg ms ⁻¹
Final momentum = mv= 100 × 8 = 800 kg ms⁻¹
Force exerted on the object, 𝐹 = 𝑚𝑣−𝑚𝑢/𝑡
= m(v-u)/t = 800-500/6 = 300/6 = 50N
Initial momentum of the object is 500 kg ms⁻¹
.
Final momentum of the object is 800 kg ms-1.
Force exerted on the object is 50 N.
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An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Mass of the object, m1 = 1 kg Velocity of the object before collision, v₁ = 10 m/s Mass of the stationary wooden block, m₂ = 5 kg Velocity of the wooden block before collision, v₂= 0 m/s ∴ Total momentum before collision = m₁ v₁ + m₂ v₂ = 1 (10) + 5 (0) = 10 kg m s ⁻¹ It is given that after collisioRead more
Mass of the object, m1 = 1 kg
Velocity of the object before collision, v₁ = 10 m/s
Mass of the stationary wooden block, m₂ = 5 kg
Velocity of the wooden block before collision, v₂= 0 m/s
∴ Total momentum before collision = m₁ v₁ + m₂ v₂
= 1 (10) + 5 (0) = 10 kg m s ⁻¹
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁ v₁ + m₂ v₂ = (m1+ m₂) v
1 (10) + 5 (0) = (1 + 5) v
V=10/6 = 5/3 m/s
The total momentum after collision = 10 and velocity of combined object is 5/3
m/s.
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A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude
Now, it is given that the bullet is travelling with a velocity of 150 m/s. Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s Final velocity, v = 0 (since the bullet finally comes to rest) Time taken to come to rest, t= 0.03 s According to the first equation of motiRead more
Now, it is given that the bullet is travelling with a velocity of 150 m/s.
Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v = u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)
A = -150/0.03 = – 5000 m/s²
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v²= u² + 2as
0 = (150)² + 2 ( – 5000) s
S = – (150)²/-2(5000) = 22500/10000 = 2.25m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass x Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s²
F = ma = 0.01×5000 = 50 N
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A hockey ball of mass 200 g travelling at 10 m s–1 is struck bya hockey stick so as to return it along its original path with a velocity at 5 m s–1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Mass of the hockey ball, m= 200 g = 0.2 kg Hockey ball travels with velocity, v₁= 10 m/s Initial momentum = mv₁ Hockey ball travels in the opposite direction with velocity, v₂= -5 m/s Final momentum = mv₂ Change in momentum = mv₁- mv₂= 0.2 [10 - (-5)] = 0.2 (15) = 3 kg m s⁻¹ Hence, the change in momRead more
Mass of the hockey ball, m= 200 g = 0.2 kg
Hockey ball travels with velocity, v₁= 10 m/s
Initial momentum = mv₁
Hockey ball travels in the opposite direction with velocity, v₂= -5 m/s
Final momentum = mv₂
Change in momentum = mv₁- mv₂= 0.2 [10 – (-5)] = 0.2 (15) = 3 kg m s⁻¹
Hence, the change in momentum of the hockey ball is 3 kg m s⁻¹.
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According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
The truck has a large mass. Therefore, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in onRead more
The truck has a large mass. Therefore, the static friction between the truck and
the road is also very high. To move the car, one has to apply a force more than
the static friction. Therefore, when someone pushes the truck and the truck does
not move, then it can be said that the applied force in one direction is cancelled
out by the frictional force of equal amount acting in the opposite direction.
Therefore, the student is right in justifying that the two opposite and equal cancel
each other.
For more answers visit to website:
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Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Mass of one of the objects, m₁ = 1.5 kg Mass of the other object, m₂ = 1.5 kg Velocity of m1 before collision, v₁ = 2.5 m/s Velocity of m₂, moving in opposite direction before collision, v₂ = -2.5 m/s (Negative sign arises because mass m₂ is moving in an opposite direction) After collision, the twoRead more
Mass of one of the objects, m₁ = 1.5 kg
Mass of the other object, m₂ = 1.5 kg
Velocity of m1 before collision, v₁ = 2.5 m/s
Velocity of m₂, moving in opposite direction before collision, v₂ = -2.5 m/s
(Negative sign arises because mass m₂ is moving in an opposite direction)
After collision, the two objects stick together.
Total mass of the combined object = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁ + v₁ + m₂ v₂= (m₁ + m₂) v
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5) v
3.75 – 3.75 = 3 v
v = 0
Hence, the velocity of the combined object after collision is 0 m/s.
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Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
A force of 200 N is applied in the forward direction. Thus, from Newton's third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the fictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet. For more answerRead more
A force of 200 N is applied in the forward direction. Thus, from Newton’s third
law of motion, an equal amount of force will act in the opposite direction. This
opposite force is the fictional force exerted on the cabinet. Hence, a frictional
force of 200 N is exerted on the cabinet.
For more answers visit to website:
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What is the momentum of an object of mass m, moving with a velocity v?
(d) mv Mass of the object = m Velocity = v Momentum = Mass × Velocity Momentum = mv
(d) mv
See lessMass of the object = m
Velocity = v
Momentum = Mass × Velocity
Momentum = mv