Hi I see another student actually posted this question... but got no answer... is someone able to look at it? see below -
a bullet is fired at and passes a piece of target paper suspended by a massless string. The bullet has a mass of m, a speed v before the collision with the target, and a speed (0.466)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.453)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of mass m of the bullet and speed v of the bullet before the collision. Express answer t0 3 decimals.
To solve this problem, we can use the principle of conservation of momentum and energy. Here's how:
Step 1: Conservation of momentum before and after the collision.
According to the principle of conservation of momentum, the total momentum of the system remains constant. In this case, the system consists of the bullet and the target.
Before the collision:
The momentum of the bullet before the collision (p_bullet) is given by:
p_bullet = m_bullet * v_bullet
After passing through the target, the bullet's momentum becomes zero since it stops. Therefore, the target's momentum after the collision will be equal in magnitude and opposite in direction to the bullet's momentum before the collision:
p_target = -p_bullet
Step 2: Conservation of energy during the collision.
During the inelastic collision, energy is lost. The amount of energy lost is given as a fraction [(0.453)KE_b BC] of the kinetic energy (KE_bullet) of the bullet before the collision.
The kinetic energy of the bullet before the collision (KE_bullet) is given by:
KE_bullet = (1/2) * m_bullet * v_bullet^2
The energy lost during the collision (ΔE) is given by the fraction multiplied by the kinetic energy of the bullet:
ΔE = (0.453) * KE_bullet
Step 3: Determining the mass of the target (M).
We can equate the kinetic energy lost during the collision to the change in kinetic energy of the system (which is equal to the kinetic energy of the target after the collision) to solve for the mass of the target.
The kinetic energy of the target after the collision (KE_target) is given by:
KE_target = (1/2) * M * V^2
We can rewrite the equation as follows:
ΔE = (1/2) * M * V^2 - 0
Substituting the equation for ΔE from step 2, we get:
(0.453) * KE_bullet = (1/2) * M * V^2
Step 4: Solving for the mass of the target (M).
Using the equation from step 3, we can solve for M:
M = (2 * (0.453) * m_bullet * v_bullet^2) / V^2
Step 5: Determining the speed of the target (V).
To find the speed of the target after the collision (V), we need another equation. We can use the principle of conservation of momentum again.
Considering the x-axis as the direction of motion:
Before the collision, the total momentum is zero since the bullet and target are stationary.
After the collision, the total momentum is also zero since the bullet stops and the target starts moving.
Using this principle, we can write another equation:
(m_bullet * v_bullet) + (M * V) = 0
From this equation, we can solve for V:
V = -(m_bullet * v_bullet) / M
Step 6: Plug in values to get the final answer.
To express the solution in terms of the mass m of the bullet and speed v of the bullet before the collision, substitute the values:
M = (2 * (0.453) * m * v^2) / V^2
V = -(m * v) / M
To determine the mass M of the target and the speed V of the target after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
Let's break down the problem step by step:
1. Conservation of momentum:
Before the collision, the total momentum of the system (bullet + target) is given by the equation:
mv = mVb + MVt, where Vb is the velocity of the bullet after passing through the target, and Vt is the velocity of the target after the collision.
2. Conservation of kinetic energy:
Before the collision, the kinetic energy of the system is given by the equation:
(1/2)mv^2 = (1/2)m(Vb)^2 + (1/2)MVt^2
3. Energy loss during collision:
The problem states that during the collision, the amount of energy lost is equal to a fraction [(0.453)KEb BC] of the kinetic energy of the bullet before the collision. KEb denotes the kinetic energy of the bullet before the collision.
4. Using the given information:
The problem states that the velocity of the bullet after passing through the target is (0.466)v. So, Vb = (0.466)v.
Now, let's put all the information together:
From step 1, we have:
mv = m(0.466)v + MVt
From step 2, we have:
(1/2)mv^2 = (1/2)m((0.466)v)^2 + (1/2)MVt^2
From step 3, we know that the energy loss during the collision is equal to [(0.453)KEb BC] times the kinetic energy of the bullet before the collision. So, energy loss = (0.453)(1/2)mv^2.
Now, we can substitute the values and solve for M and Vt:
(0.534)mv = MVt (step 1)
(1/2)mv^2 - (0.105)mv^2 = (1/2)MVt^2 (step 2)
Now, divide both equations by mv to get:
(0.534) = Vt (from step 1)
(1/2)m - (0.105)m = (1/2)MVt (from step 2)
Simplify the second equation:
(1/2)m - (0.105)m = (1/2)M(0.534)
Solve for M:
M = [(1/2)m - (0.105)m] / [(1/2)(0.534)]
Now, you can substitute the given values for m, v, and solve the equation to find the value of M. Similarly, substitute the value of M in step 1 to find Vt. Remember to express the answer to three decimal places.
Good lord, talk about busy work. I think I would look for another instructor, this one likes to not teach, but demonstrate how clever they are.
First, you know the speed of the bullet the bullet in terms of initial v.
you also know Ke as a fraction..
KE: finalKE+energy lost=inital KE
1/2 m (.466v)^2+.453*1/2 m v^2=1/2mv^2
somehow, I see inconsistency here.