An archer shoots an arrow toward a 300 g target that is sliding in her direction at a speed of 2.15 m/s on a smooth, slippery surface. The 22.5 g arrow is shot with a speed of 38.0 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?
THis can be found using the conservation of momentum.
Pix + Pix = Pfx + Pfx
Where P = mv
The mass of the arrow is .3 kg, the target is .0225kg. The initial velocity of the target is -2.15 m/s and the initial velocity of the arrow is 38.0m/s. If the target is stopped, the final velocity is 0 m/s. SO:
(.3kg)(38.0m/s) + (.0225kg)(-2.15m/s) = (.3kg)(Vf) + (.0225kg)(0m/s)
[(11.4Ns) + (-.048375Ns)]/(.3kg)=Vf
I'll let you figure it out from there. ;)
This is wrong. The 0m/s should be with the mass of the target, not of the arrow.
To find the speed of the arrow after passing through the target, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity (p = m * v). In this case, the momentum before the collision is the sum of the momenta of the arrow and the target.
Let's denote:
m1 = mass of the arrow (22.5 g = 0.0225 kg)
m2 = mass of the target (300 g = 0.3 kg)
v1 = initial velocity of the arrow (38.0 m/s)
v2 = initial velocity of the target (2.15 m/s)
v_f = final velocity of the arrow after passing through the target (to be found)
The initial total momentum is:
p_initial = (m1 * v1) + (m2 * v2)
The final total momentum is:
p_final = (m1 * v_f) + (m2 * 0) (since the target is stopped by the impact)
According to the conservation of momentum, we have p_initial = p_final.
Substituting the given values into the equation, we get:
(m1 * v1) + (m2 * v2) = (m1 * v_f) + (m2 * 0)
Now, let's solve for v_f:
(m1 * v1) + (m2 * v2) = (m1 * v_f) + 0
m1 * v1 + m2 * v2 = m1 * v_f
0.0225 kg * 38.0 m/s + 0.3 kg * 2.15 m/s = 0.0225 kg * v_f
0.855 kg m/s + 0.645 kg m/s = 0.0225 kg * v_f
1.5 kg m/s = 0.0225 kg * v_f
To find v_f, we divide both sides of the equation by 0.0225 kg:
v_f = (1.5 kg m/s) / 0.0225 kg
v_f ≈ 66.67 m/s
Therefore, the speed of the arrow after passing through the target is approximately 66.67 m/s.
To find the speed of the arrow after passing through the target, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision, the momentum of the target is given by:
Momentum_target = mass_target × velocity_target
= 0.300 kg × 2.15 m/s
= 0.645 kg⋅m/s
Similarly, the momentum of the arrow before the collision is given by:
Momentum_arrow = mass_arrow × velocity_arrow
= 0.0225 kg × 38.0 m/s
= 0.855 kg⋅m/s
Since the arrow passes through the target and stops, their final velocities are the same. Let's call this final velocity "v":
After the collision, the momentum of the arrow and the target combined is given by:
Momentum_combined = (mass_target + mass_arrow) × v
By applying the conservation of momentum, we equate the total momentum before and after the collision:
Momentum_target + Momentum_arrow = Momentum_combined
Substituting the values we calculated earlier, we have:
0.645 kg⋅m/s + 0.855 kg⋅m/s = (0.300 kg + 0.0225 kg) × v
Simplifying the equation further:
1.50 kg⋅m/s = 0.3225 kg × v
Now, we can solve for the final velocity "v":
v = (1.50 kg⋅m/s) / 0.3225 kg
v ≈ 4.65 m/s
Therefore, the speed of the arrow after passing through the target is approximately 4.65 m/s.