A 5.00-g bullet is fired into a 500-g block of wood suspended as a ballistic pendulum.
The combined mass swings up to a height of 10.00 cm. What was the magnitude of the
momentum of the combined mass immediately after the collision?
The Ans is 0.707 kg m/sec
I do not know how to go get this answer
Energy is conserved during the upward swing. Therefore
(M+m)V = (M+m)gH
and
V = gH = 0.98 m/s
is the block velocity just after the bullet (mass m) is embedded inside the block (mass M). The momentum at that time is
(M+m)V = (0.505 kg)*V = 0.495 kg m/s
I do not agree with the 0.707 answer.
To find the magnitude of momentum of the combined mass immediately after the collision, we can use the principles of conservation of momentum.
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, provided that no external forces are acting on the system.
In this case, the bullet is fired into the block of wood, resulting in a collision. The bullet and the block of wood then move together as a combined mass.
First, let's find the initial momentum of the bullet. We know the mass of the bullet is 5.00 g, which is equal to 0.005 kg (since 1 g = 0.001 kg). We also need to know the velocity of the bullet before the collision. Let's assume it is v.
The momentum of the bullet is given by:
Initial momentum of the bullet = mass x velocity = 0.005 kg x v = 0.005v kg m/s
Next, let's find the initial momentum of the block of wood. The mass of the block of wood is given as 500 g, which is equal to 0.500 kg. The initial velocity of the block of wood is zero since it is at rest before the collision.
Therefore, the initial momentum of the block of wood is:
Initial momentum of the block of wood = mass x velocity = 0.500 kg x 0 = 0 kg m/s
Since momentum is conserved, the total initial momentum before the collision is equal to the total final momentum after the collision.
Total initial momentum = Total final momentum
Therefore,
0.005v + 0 = Total final momentum
After the collision, the combined mass swings up to a height of 10.00 cm. This indicates that the initial kinetic energy (associated with the motion of the bullet and block of wood) is converted into potential energy (due to the height reached by the combined mass).
Since momentum is directly proportional to velocity, and kinetic energy is directly proportional to the square of velocity, we can equate the initial kinetic energy to the final potential energy.
Initial kinetic energy = Final potential energy
Therefore,
0.5 x (0.005v)^2 = m x g x h
Where m is the mass of the combined mass and equals 0.505 kg (0.500 kg for the block of wood + 0.005 kg for the bullet), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height reached (0.10 m since 10.00 cm = 0.10 m).
Simplifying the equation, we get:
0.0000125v^2 = 0.505 x 9.8 x 0.10
0.0000125v^2 = 0.04951
Divide through by 0.0000125 to solve for v^2:
v^2 = 0.04951 / 0.0000125
v^2 = 3960.8
Taking the square root of both sides, we find v:
v = √3960.8 ≈ 62.97 m/s
Substituting this value of v back into the equation for initial momentum of the bullet:
Initial momentum of the bullet = 0.005v = 0.005 x 62.97 ≈ 0.3149 kg m/s
Since momentum is conserved, the magnitude of the momentum of the combined mass immediately after the collision is equal to the initial momentum of the bullet:
Magnitude of momentum of the combined mass = 0.3149 kg m/s
However, the provided answer is 0.707 kg m/s. It is possible that there was a calculation error or a mistake in the given answer. Please double-check your calculations or refer to the original source for clarification.