A student flings a 23g ball of putty at a 225g cart sitting on a slanted air track that is 1.5m long. The track is slanted at an angle of 25deg with the horizontal. If the putty is traveling at 4.2m/s when it hits the cart, will the cart reach the end of the track before it stops and slides back down? Support your answer with calculations
Note that the putty and the card will move together after they gained momentum, meaning the putty will stick on the cart. Thus first let's calculate the velocity of the putty and the cart together after the momentum:
Lets mp, mc, vp, and vc represent the mass of the putty, mass of the cart, speed of the putty, and speed of the cart. And Vpc represent the speed of the cart and the putty together:
Based on the principle of conservation of momentum, we have:
mpvp + mcvc = mpv'p + mcv'c
since the cart was stationary, vc = 0; and since the putty and the cart moves together after they gained momentum, the V'pc represent their velocity after momentum. Then we will get the following equation:
mpvp = (mp + mc) V'pc
V'pc = mpvp / (mp + mc)
V'pc = (0.023 Kg * 4.2m/s) / (0.023 kg + 0.225 Kg)
V'pc = 0.3895m/s
and from the conservation of energy we have,
Ek = Eg
1/2*mcp*v^2 = mcp * g * h, where mcp represent the mass of the putty and the cart together
1/2*v^2 = g*h , after we devide both sides of the above equation by mcp
then, h = v^2/2g
h = (0.3895)^2/(2*9.81)
h = 7.732 * 10^-3
and we know that ;
sin25 = h/hypotenuse, where hypotenuse is the distance the cart and the putty travels
hypotenuse = h / sin25
hypotenuse =( 7.32 * 10^-3) / sin25
hypotenuse = 0.0182964849
hypotenuse = 0.018m
the distance that the cart and the putty travels together is equal to the hypotenuse, which is 0.018m Thus, the cart don't make the end of the track, which his 1.5 m long, before it stops.
I tried solving this by separating the components into the vertical and horizontal, before and after collision. but i only obtain velocity and the deg, however the questions asks for the distance...
The last calculation: 0.007732/sin25=0.02020 not 0.018
To determine whether the cart will reach the end of the track before it stops and slides back down, we need to calculate the initial velocity of the cart after being struck by the putty ball.
First, let's determine the horizontal component of the putty ball's velocity. We can use the equation:
Vx = V * cos(θ)
Where:
Vx is the horizontal component of velocity
V is the initial velocity of the putty ball
θ is the angle of the track with the horizontal (25 degrees)
Plugging in the given values:
Vx = 4.2 m/s * cos(25°)
Calculating it:
Vx = 4.2 m/s * 0.9063
Vx ≈ 3.8125 m/s (rounded to four decimal places)
Next, let's calculate the horizontal distance covered by the cart, using the equation:
d = Vx * t
Where:
d is the distance covered by the cart
Vx is the initial horizontal velocity of the cart (3.8125 m/s)
t is the time taken to reach the end of the track
We can find the time taken using the kinematic equation:
d = V0 * t + (1/2) * a * t^2
Since the acceleration along the horizontal direction is zero (no external forces), the equation simplifies to:
d = V0 * t
Rearranging the equation to solve for time:
t = d / V0
Plugging in the given values:
t = 1.5 m / 3.8125 m/s
Calculating it:
t ≈ 0.39368 s (rounded to five decimal places)
Now, let's calculate the vertical distance covered by the cart while sliding down the incline. We can use the equation:
D = h * sin(θ)
Where:
D is the vertical distance covered by the cart
h is the height of the inclined track
Since the height is not given, we assume it to be negligible or equal to zero. Therefore:
D = 0
Since the vertical distance covered by the cart is zero, it means that the cart will remain on the track. Hence, the cart will reach the end of the track before it stops and slides back down.
Therefore, the cart will reach the end of the track before it stops and slides back down.
Wouldn't it matter the direction the putty was traveling?
I am not going to work this for you. I will be happy to critique your thinknig.