if 15g bullet traveling at 300ms-¹ strike and is absorb by a 75kg object find the speed with which the object moved off.
conserve momentum
15*300 = (15+75000)v
Well, let's put on our physics hats and crunch some numbers! If a 15g (or 0.015kg) bullet traveling at 300 m/s strikes and is absorbed by a 75kg object, we can use the principle of conservation of momentum to find the speed with which the object moves off.
The momentum of the bullet (before impact) can be calculated by multiplying its mass by its velocity, which is 0.015kg × 300 m/s = 4.5 kg⋅m/s.
Since momentum is conserved in this case, we can set the momentum of the object (after impact) equal to the momentum of the bullet. Therefore, the momentum of the object is also 4.5 kg⋅m/s.
Now, to find the speed at which the object moves off, we need to divide its momentum by its mass. So, 4.5 kg⋅m/s ÷ 75 kg gives us a speed of 0.06 m/s.
Therefore, the object moved off with a speed of approximately 0.06 m/s. However, it's important to note that this answer assumes a perfectly elastic collision, where no energy is lost. In reality, there might be some energy dissipation during the collision, so keep that in mind!
To find the speed with which the object moves off after being struck by a 15g bullet traveling at 300 m/s, you can use the principle of conservation of momentum.
The momentum before the collision (bullet) is equal to the momentum after the collision (object).
Momentum = mass x velocity
First, convert the mass of the bullet from grams to kilograms:
Mass of bullet = 15g = 15/1000 kg = 0.015 kg
Now, we can calculate the momentum before the collision:
Momentum of bullet before = Mass of bullet x Velocity of bullet
Momentum of bullet before = 0.015 kg x 300 m/s = 4.5 kg m/s
Since the object absorbs the bullet, the final momentum after the collision will be zero (due to the bullet's initial momentum cancelling out with the object's final momentum).
Momentum of object after = 0 kg m/s
Applying the conservation of momentum equation:
Momentum of bullet before = Momentum of object after
4.5 kg m/s = Mass of object x Velocity of object after
Since the mass of the object is 75 kg, we can rearrange the equation to solve for the velocity of the object after the collision:
Velocity of object after = Momentum of bullet before / Mass of object
Velocity of object after = 4.5 kg m/s / 75 kg
Velocity of object after = 0.06 m/s
Therefore, the object moves off with a speed of 0.06 m/s after being struck by the bullet.
To find the speed with which the object moves off after absorbing the bullet, we can use the principle of conservation of momentum.
The momentum before the collision is equal to the momentum after the collision. Mathematically, we can express this as:
Momentum before = Momentum after
The momentum of an object is given by the product of its mass and velocity. Let's assume the initial velocity of the object is zero.
Momentum before = mass of bullet × velocity of bullet
Momentum after = mass of object × velocity of object
Given:
Mass of bullet (m1) = 15 g = 15/1000 kg = 0.015 kg
Velocity of bullet (v1) = 300 m/s
Mass of object (m2) = 75 kg
Velocity of object (v2) = ?
Using the conservation of momentum equation, we get:
m1v1 = m2v2
Substituting the given values, we have:
(0.015 kg)(300 m/s) = (75 kg)(v2)
Now we can solve for v2:
0.015 × 300 = 75 × v2
4.5 = 75 × v2
v2 = 4.5 / 75
v2 = 0.06 m/s
Therefore, the object moves off with a speed of 0.06 m/s after absorbing the bullet.