A m = 1.4 kg object moving at v = 15 m/s collides with a stationary 2.0 kg object. If the collision is perfectly inelastic, how far along the inclined plane (37degrees) will the combined system travel? Neglect friction.
I have never know what a perfectly inelastic collision is. I do know what an inelastic collision is: Energy is NOT conserved. Ask your teacher this question, see if they know.
Momentum applies:
Now on this statement, conservation ofm momentum applies
momentum before=momentum after
along the path of the moving object..
1.4*15m/s+2*0=(6kg*V) assuming the objects are stuck together, the problem did not state that.
Now, you can consider energy:
initialKE after collision= change PEnergy
1/2 (6)V^2= (6)g height
where distancealongplane=hSin37
oops...
distancealongplane=h/sin37
where are you getting 6kg from?
To find the distance traveled by the combined system after the collision, we need to analyze the conservation of momentum and the conservation of energy.
Step 1: Conservation of Momentum
In an inelastic collision, the two objects stick together after the collision. This means that the final velocity of the combined system can be found using the conservation of momentum equation:
m1 * v1 + m2 * v2 = (m1 + m2) * vf
Where:
m1 = mass of the first object (1.4 kg)
v1 = initial velocity of the first object (15 m/s)
m2 = mass of the second object (2.0 kg)
v2 = initial velocity of the second object (0 m/s)
vf = final velocity of the combined system
Substituting the given values into the equation:
(1.4 kg) * (15 m/s) + (2.0 kg) * (0 m/s) = (1.4 kg + 2.0 kg) * vf
Step 2: Solve for Final Velocity (vf)
Simplifying the equation:
(21 kg∙m/s) = (3.4 kg) * vf
Divide both sides by 3.4 kg:
vf = (21 kg∙m/s) / (3.4 kg)
vf ≈ 6.1765 m/s
Step 3: Conservation of Energy
To find the distance traveled by the combined system, we will use the conservation of energy equation. The initial kinetic energy (KE_initial) of the system before the collision is equal to the final potential energy (PE_final) of the system after it travels up the inclined plane.
KE_initial = PE_final
The initial kinetic energy is given by:
KE_initial = 0.5 * m_total * v_final^2
The final potential energy is given by:
PE_final = m_total * g * h
Where:
m_total = total mass of the combined system (1.4 kg + 2.0 kg)
v_final = final velocity of the combined system (6.1765 m/s)
g = acceleration due to gravity (9.8 m/s^2)
h = height traveled up the inclined plane
Step 4: Solve for Height (h)
Setting the initial kinetic energy equal to the final potential energy:
0.5 * m_total * v_final^2 = m_total * g * h
Divide both sides by m_total:
0.5 * v_final^2 = g * h
Solve for h:
h = (0.5 * v_final^2) / g
Substituting the given values into the equation:
h = (0.5 * (6.1765 m/s)^2) / (9.8 m/s^2)
h ≈ 1.98 m
Therefore, the combined system will travel approximately 1.98 meters along the inclined plane.