Use the equations below:
P=mv
Ek=1/2mv^2
4) A wooden target of mass 2.00kg is suspended from a branch so that it swings freely. The target is hit by an arrow of mass 0.10kg travelling horizontally and the arrow embeds itself in the centre of the target. The target swings upwards, with its centre of mass rising by 0.60m. Use 9.81 as the acceleration for gravity.
A) Calculate the gain in gravitational energy by the target and arrow combination.
B) Calculate the velocity of the arrow and target immediately after the impact.
C) Calculate the velocity of the arrow immediately before the impact.
D) Determine if the collision is elastic or inelastic and why.
12 marks
please answer
This one is a really tricky one so please write out in full, and no symbols please as i dont understand when you do that just full words
To solve this problem, we will use the equations P = mv and Ek = 1/2mv^2.
A) To calculate the gain in gravitational energy by the target and arrow combination, we need to find the change in gravitational potential energy (ΔPE). The formula for gravitational potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity (9.81 m/s^2), and h is the height.
In this case, the target swings upwards by a height of 0.60 m. The change in gravitational potential energy is given by:
ΔPE = mgh = (2.00 kg)(9.81 m/s^2)(0.60 m) = 11.77 J
So, the gain in gravitational energy by the target and arrow combination is 11.77 J.
B) To calculate the velocity of the arrow and target immediately after the impact, we will use the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
The momentum before the collision is given by the equation P = mv. The arrow is traveling horizontally, so its momentum before the collision is (0.10 kg)(v), where v is its velocity. The target is initially at rest, so its momentum before the collision is 0. After the collision, the arrow and target move together, so their momentum after the collision is (2.00 kg + 0.10 kg)(v'), where v' is their common velocity.
Using the law of conservation of momentum, we can equate the initial momentum to the final momentum:
(0.10 kg)(v) = (2.10 kg)(v')
Simplifying the equation, we find:
v' = (0.10 kg)(v) / (2.10 kg) = 0.0476v
Therefore, the velocity of the arrow and target immediately after the impact is 0.0476 times the velocity of the arrow before the impact.
C) To calculate the velocity of the arrow immediately before the impact, we need to use the principle of conservation of mechanical energy, which states that the total mechanical energy before the collision is equal to the total mechanical energy after the collision.
The initial mechanical energy is the kinetic energy of the arrow before the collision, given by the equation Ek = 1/2mv^2. Plugging in the values, we have:
Ek = (1/2)(0.10 kg)(v^2)
The final mechanical energy is the sum of the kinetic energy of the arrow and the gain in gravitational energy by the target and arrow combination. So, we can write:
Ek' + ΔPE = (1/2)(2.10 kg)(v')^2 + 11.77 J
Simplifying the equation, we find:
(1/2)(0.10 kg)(v^2) + 11.77 J = (1/2)(2.10 kg)(0.0476v)^2 + 11.77 J
Now we can solve this equation to find the value of v.
D) To determine if the collision is elastic or inelastic, we need to calculate the coefficient of restitution (e) using the formula:
e = (v1' - v2') / (v1 - v2)
In this case, since the target and arrow move together after the collision, their velocities are the same (v'). Therefore, substituting the values, the equation becomes:
e = (v' - v') / (v - v2)
Simplifying the equation, we find:
e = 0 / (v - v2) = 0
Since the coefficient of restitution (e) is 0, the collision is considered perfectly inelastic. This means that the objects stick together after the collision and there is a loss of kinetic energy.
By solving the equations for parts B and C, you should be able to obtain the numerical values for the velocities.