A system of inertia 0.39 kg consists of a spring gun attached to a cart. The system is at rest on a horizontal low-friction track. A 0.050-kg projectile is loaded into the gun, then launched at an angle of 40∘ with respect to the horizontal plane.
With what speed does the cart recoil if the projectile rises 2.3 m at its maximum height?
DO YOUR HOMEWORK!!
0.579 m/s
1.2 m/s. Don't know how but that what it says. Marvin, provide an answer or be quiet.
OMG YOU KNOW MARVIN?? AT THE U???
To determine the recoil speed of the cart, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, as long as no external forces act on the system.
In this case, the initial momentum is zero because the system is at rest. The final momentum will be the momentum of the cart after the projectile is launched.
Let's calculate the momentum of the projectile first. The momentum (p) of an object is given by the formula:
p = m * v
Where:
m is the mass of the object, and
v is the velocity of the object.
In this case, the mass (m) of the projectile is 0.050 kg. We need to find the velocity (v) of the projectile.
To do this, we first need to determine the initial velocity of the projectile. This can be calculated using the maximum height reached by the projectile.
The vertical motion of the projectile is governed by the equation:
y = (v^2 * sin^2θ) / (2g)
Where:
y is the maximum height reached by the projectile,
v is the initial velocity of the projectile,
θ is the launch angle with respect to the horizontal plane, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the maximum height (y) is given as 2.3 m and the launch angle (θ) is 40°.
Plugging the values into the equation, we can solve for v:
2.3 m = (v^2 * sin^2(40°)) / (2 * 9.8 m/s^2)
Now, we can solve for v:
v^2 = (2.3 m * 2 * 9.8 m/s^2) / sin^2(40°)
v^2 ≈ 68.20 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 8.26 m/s
Now that we have the velocity of the projectile, we can use the principle of conservation of linear momentum to determine the recoil speed of the cart.
According to the principle, the total momentum before the projectile is launched is zero. Therefore, the momentum of the cart after the projectile is launched will be equal in magnitude but opposite in direction to the momentum of the projectile.
The momentum of the cart (m_cart * v_cart) will be equal to the momentum of the projectile (0.050 kg * (-8.26 m/s)).
We can rearrange the formula to solve for v_cart:
v_cart = (-m_projectile * v_projectile) / m_cart
Given that the mass of the cart (m_cart) is 0.39 kg, the mass of the projectile (m_projectile) is 0.050 kg, and the velocity of the projectile (v_projectile) is -8.26 m/s, we can substitute these values into the equation:
v_cart = (-0.050 kg * -8.26 m/s) / 0.39 kg
v_cart ≈ 1.07 m/s
Therefore, the cart recoils with a speed of approximately 1.07 m/s.