A bullet of mass 45 g is fired at a speed of 220 m/s into a 5.0-kg sandbag hanging from a string from the ceiling. The sandbag absorbs the bullet and begins to swing. To what maximum vertical height will it rise? Hint: Energy is not conserved as the bullet enters the bag but is conserved after the bullet comes to rest in the bag and the bag is swinging upward
mv=(m+M)u
u=mv/(m+M)=0.045•220/(0.045+5)=1.96 m/s
(m+M)u²/2=(m+M)gh
h= u²/2g=1.96²/2•9.8=0.196 m
v1'=((m1-m2)/(m1+m2))v1 for the bullet =176ms^-1
And for the sandbag at rest
v2'=((2m1)/(m1+m2))v1 =396ms^-1
The bullet is conserved after the bullet comes to rest in the bag.
Hence Mgh= 1/2mv^2.
I got stuck here what is going to be the value of M, is it sum of the two masses. and value of v as well.
And i hope the above is collect.
Thanks
To determine the maximum vertical height the sandbag will rise, we need to consider the conservation of energy.
Step 1: Calculate the kinetic energy of the bullet initially.
The kinetic energy of an object is given by the equation: KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
Given:
Mass of the bullet (m) = 45 g = 0.045 kg
Initial speed of the bullet (v) = 220 m/s
Using the formula:
KE = 1/2 * 0.045 kg * (220 m/s)^2
KE = 0.5 * 0.045 * 220^2
KE = 539.7 J
Step 2: Determine the total energy of the system after the bullet comes to rest in the bag.
After the bullet comes to rest in the bag, the total energy of the system is conserved. Since the bullet is absorbed and the bag starts swinging upwards, the energy is converted into potential energy.
Step 3: Calculate potential energy.
The potential energy of an object is given by the equation: PE = m * g * h, where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.
Given:
Mass of the sandbag (m) = 5.0 kg
Acceleration due to gravity (g) = 9.8 m/s^2
We need to find the maximum height (h) to which the sandbag rises.
Using the formula:
PE = m * g * h
539.7 J = 5.0 kg * 9.8 m/s^2 * h
Simplifying the equation:
h = 539.7 J / (5.0 kg * 9.8 m/s^2)
h = 11.0 meters
Therefore, the sandbag will rise to a maximum vertical height of 11.0 meters.
To find the maximum vertical height the sandbag will rise after the bullet is absorbed, we need to apply the principle of conservation of mechanical energy.
The initial mechanical energy of the system consists of the kinetic energy (KE) of the bullet, and the potential energy (PE) and kinetic energy of the hanging sandbag.
We know the initial kinetic energy of the bullet is given by:
KE_bullet = (1/2) * (mass_bullet) * (velocity_bullet)^2
Converting the mass of the bullet from grams to kilograms:
mass_bullet = 45 g = 0.045 kg
Plugging in the values and solving for KE_bullet:
KE_bullet = (1/2) * 0.045 kg * (220 m/s)^2 = 539.1 J
Since the bullet is absorbed by the sandbag and comes to rest, all of its initial kinetic energy is transferred to the sandbag. Therefore, the initial kinetic energy of the sandbag is also 539.1 J.
As the sandbag rises to its maximum vertical height, its kinetic energy will convert entirely to potential energy. Thus, at the highest point, the kinetic energy of the sandbag becomes zero, and all the initial energy is transformed into potential energy.
Setting the final kinetic energy of the sandbag to zero, and expressing potential energy in terms of height:
PE_max = KE_bullet + KE_sandbag = m_sandbag * g * h_max
Where m_sandbag is the mass of the sandbag, g is the gravitational acceleration, and h_max is the maximum vertical height.
Given:
mass_sandbag = 5.0 kg
g = 9.8 m/s^2
Solving for h_max:
h_max = (KE_bullet + KE_sandbag) / (m_sandbag * g)
Substituting in the known values:
h_max = (539.1 J + 0) / (5.0 kg * 9.8 m/s^2) = 10.98 m
Therefore, the maximum vertical height the sandbag will rise is approximately 10.98 meters.