As shown in the figure 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.
btw obviously I did not include a pic. THANKS
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 denote the mass of the bullet as "m", the velocity of the bullet before the collision as "v", the velocity of the bullet after passing through the target as (0.466)v, the mass of the target as "M", and the velocity of the target after the collision as "V".
Conservation of momentum tells us that the total momentum before the collision is equal to the total momentum after the collision:
(m)(v) = (m + M)(0.466)v + MV
Now, let's consider conservation of kinetic energy. We're told that the amount of energy lost during the collision is equal to a fraction [(0.453)KEb BC] of the initial kinetic energy of the bullet. Since the initial kinetic energy of the bullet is 0.5mv^2, the energy lost is (0.453)(0.5mv^2). Therefore, the kinetic energy after the collision is equal to 0.5mv^2 - (0.453)(0.5mv^2), which simplifies to (0.547)(0.5mv^2).
Conservation of kinetic energy tells us that the total kinetic energy before the collision is equal to the total kinetic energy after the collision:
0.5mv^2 = 0.5(m + M)(0.466)v^2 + 0.5MV^2
Now we have two equations:
1) (m)(v) = (m + M)(0.466)v + MV
2) 0.5mv^2 = 0.5(m + M)(0.466)v^2 + 0.5MV^2
We can solve these two equations simultaneously to find the values of M and V.
Step 1: Expand equation 1:
mv = (0.466)mv + (0.466)Mv + MV
Step 2: Simplify equation 1:
0 = (0.466)mv + (0.466)Mv + MV - mv
Step 3: Collect like terms in equation 1:
0.534mv = (0.466)Mv + MV
Step 4: Divide both sides of equation 1 by v:
0.534m = 0.466M + M
Step 5: Combine like terms in equation 1:
0.534m = 1.466M
Step 6: Divide both sides of equation 1 by 1.466:
M = (0.534m) / 1.466
So, the mass M of the target is (0.534m) / 1.466.
Now, let's substitute this value of M into equation 2:
0.5mv^2 = 0.5[(m + (0.534m) / 1.466)](0.466)v^2 + 0.5(0.534m / 1.466)V^2
Step 7: Simplify equation 2:
0.5mv^2 = 0.5(1.534m)(0.466)v^2 + 0.5(0.534/1.466)mV^2
Step 8: Divide both sides of equation 2 by 0.5mv^2:
1 = (1.534)(0.466) + (0.534/1.466)(V^2/v^2)
Step 9: Simplify equation 2 further:
1 = 0.714 (approximately) + (0.364)(V^2/v^2)
Step 10: Subtract 0.714 from both sides of equation 2:
0.286 = (0.364)(V^2/v^2)
Step 11: Divide both sides of equation 2 by 0.364:
V^2/v^2 = 0.786
Step 12: Take the square root of both sides of equation 2:
V/v = √0.786
Simplifying this expression gives:
V/v ≈ 0.886
Step 13: Multiply both sides of equation 2 by v:
V ≈ (0.886)v
So, the speed of the target after the collision is approximately (0.886)v, and the mass of the target is (0.534m) / 1.466.
To solve this question, we need to use the principles of conservation of momentum and conservation of kinetic energy.
Let's define the variables:
- m: mass of the bullet
- v: speed of the bullet before the collision
- V: speed of the target after the collision
- M: mass of the target
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the bullet passes through the target, the total momentum is given by:
m * v = (m + M) * V
Now, let's consider the principle of conservation of kinetic energy. The inelastic collision implies that some kinetic energy is lost during the collision. The energy lost is equal to a fraction of the initial kinetic energy of the bullet. Therefore:
(0.453) * (1/2) * m * v^2 = (1/2) * M * V^2
Now, we have two equations. We can solve them simultaneously to find the values of M and V in terms of m and v.
First, we rearrange the momentum equation to solve for V:
V = (m * v) / (m + M)
Substituting this value of V into the kinetic energy equation:
(0.453) * (1/2) * m * v^2 = (1/2) * M * [(m * v) / (m + M)]^2
Now, we can solve this equation for M.
To solve the equation, we can multiply both sides by 2(m + M) to simplify the equation:
0.453 * m * v^2 * (m + M) = M * [(m * v)^2]
Multiplying out and simplifying:
0.453 * m * v^2 * m + 0.453 * m * v^2 * M = M * (m^2 * v^2)
Next, we isolate M terms on one side:
0.453 * m * v^2 * m - M * (m^2 * v^2) = - 0.453 * m * v^2 * M
Now, we can divide both sides by -0.453 * m * v^2 to solve for M:
M = (0.453 * m * m * v^2) / (0.453 * m * v^2 - m^2 * v^2)
Simplifying further:
M = (m^2) / (v^2 - m * v^2)
Now, we have the value of M in terms of m and v.
To find the value of V, we substitute the value of M back into the momentum equation:
V = (m * v) / (m + (m^2) / (v^2 - m * v^2))
Simplifying this further:
V = (m * v * (v^2 - m * v^2)) / (m * (v^2 - m * v^2) + m^2)
Now we have the value of V in terms of m and v.
To express the answer to 3 decimal places, replace m with the given mass of the bullet, and v with the given speed of the bullet before the collision, then calculate M and V using the derived equations.