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
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.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-10/
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
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
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.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
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.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
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
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/