A 69kg skateboarder, at rest on a frictionless surface, throws a 4.0kg rock with the velocity given by v=3(i hat)+4(jhat), where the x and y axes are both in the horizontal plane. What is the x component of the subsequent velocity of the skateboarder? WHat is the y component of th subsequent velocity of the skateboarder?
I have no idea what i am doing please help? Any helpwould be greatly appeciated.
To solve this problem, we can apply the principle of conservation of momentum. The total momentum before the rock is thrown should be equal to the total momentum after the rock is thrown.
Let's denote the skateboarder's velocity before the throw as vb, the rock's velocity as vr, and the final velocities of the skateboarder and the rock as vf and vr'.
Given:
Mass of the skateboarder (mb) = 69 kg
Mass of the rock (mr) = 4.0 kg
Velocity of the rock (vr) = 3i + 4j (in m/s)
The initial momentum of the system (skateboarder + rock) is given by:
Initial momentum (before the throw) = mb * vb + mr * vr
Since the skateboarder is at rest initially, vb = 0i + 0j, so the initial momentum simplifies to:
Initial momentum = 0i + 0j + mr * vr
After the throw, the skateboarder and the rock will have new velocities (vf and vr') respectively. The final momentum of the system (skateboarder + rock) is given by:
Final momentum (after the throw) = mb * vf + mr * vr'
The principle of conservation of momentum states that the initial momentum and the final momentum should be equal. Therefore:
Initial momentum = Final momentum
0i + 0j + mr * vr = mb * vf + mr * vr'
Now, we can equate the x and y components of the initial and final momenta separately.
X Component:
0 + mr * vr_x = mb * vf_x + mr * vr'_x (Equation 1)
Y Component:
0 + mr * vr_y = mb * vf_y + mr * vr'_y (Equation 2)
To find the x and y components of vf, we need to assume certain conditions, specifically the x and y components of vf. In the absence of any external forces or torques acting on the system, the x and y components of the velocity should be conserved.
Therefore:
vf_x = 0
vf_y = 0
Now, let's substitute these values back into Equation 1 and Equation 2:
mr * vr_x = mb * 0 + mr * vr'_x (since vf_x = 0)
mr * vr_y = mb * 0 + mr * vr'_y (since vf_y = 0)
Simplifying further, the equations become:
mr * vr_x = mr * vr'_x (Equation 3)
mr * vr_y = mr * vr'_y (Equation 4)
Since mass (m) appears on both sides of the equations, we can divide through by mr:
vr_x = vr'_x (Equation 5)
vr_y = vr'_y (Equation 6)
Equations 5 and 6 tell us that the x and y components of the rock's velocity remain unchanged after the throw. And since the skateboarder and the rock must have opposite velocities (due to conservation of momentum), we can conclude that:
The x component of the subsequent velocity of the skateboarder is -vr_x (negative of the x component of the rock's velocity)
The y component of the subsequent velocity of the skateboarder is -vr_y (negative of the y component of the rock's velocity)
Thus, the final answer is:
The x component of the skateboarder's subsequent velocity is -3 m/s.
The y component of the skateboarder's subsequent velocity is -4 m/s.