A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.5 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.42 m/s.
(a) What is the mass of the second cart? ___ g
(b) What is its speed after impact?
_____ m/s
(c) What is the speed of the two-cart center of mass? _____ m/s
The equation for the first part is
V_1Final = [(m_1-m_2)/(m_1+m_2)]*V_1Origional
I used conservation of momentum for the second part
Use
V = (m_1*v_1 + m_2*v_2)/(m_1+m_2)
for part c. I am not sure if you should use v original or finial for the velocities in the equation.
You would use the final velocities.
To solve this problem, we can use the principle of conservation of momentum and kinetic energy.
(a) To find the mass of the second cart, we need to use the conservation of momentum equation:
m1v1i + m2v2i = m1v1f + m2v2f
where m1 and m2 are the masses of the first and second carts respectively, v1i and v2i are the initial velocities, and v1f and v2f are the final velocities.
Since the second cart is initially stationary, its initial velocity (v2i) is 0 m/s. The final velocity of the first cart (v1f) is given as 0.42 m/s. The initial velocity of the first cart (v1i) is 1.5 m/s, and the final velocity of the second cart (v2f) is what we want to find.
Plugging these values into the conservation of momentum equation, we get:
(0.34 kg)(1.5 m/s) + (m2)(0 m/s) = (0.34 kg)(0.42 m/s) + (m2)(v2f)
Simplifying the equation gives us:
0.51 kg + 0 = 0.1428 kg + (m2)(v2f)
Since the carts experience an elastic collision, the total kinetic energy before and after the collision is conserved. So we can also use the equation for conservation of kinetic energy:
(1/2)(m1)(v1i)^2 + 0 = (1/2)(m1)(v1f)^2 + (1/2)(m2)(v2f)^2
Plugging in the given values, we get:
(1/2)(0.34 kg)(1.5 m/s)^2 + 0 = (1/2)(0.34 kg)(0.42 m/s)^2 + (1/2)(m2)(v2f)^2
Simplifying the equation gives us:
0.765 J = 0.03192 J + (1/2)(m2)(v2f)^2
Now we have two equations with two unknowns (m2 and v2f). We can solve these simultaneously to find the values.
(b) First, solve for m2:
0.51 kg = 0.1428 kg + (m2)(v2f)
Subtracting 0.1428 kg from both sides:
0.3672 kg = (m2)(v2f)
Now, solve for v2f using the kinetic energy equation:
0.765 J = 0.03192 J + (1/2)(m2)(v2f)^2
Subtracting 0.03192 J from both sides:
0.73308 J = (1/2)(m2)(v2f)^2
Simplifying:
1.46616 J = (m2)(v2f)^2
Now, divide the equation for m2 by the equation for v2f:
(0.3672 kg) / v2f = (1.46616 J) / (v2f)^2
Simplifying:
0.3672 kg = 1.46616 J / v2f
Now, we can equate the two expressions for m2:
0.3672 kg = 1.46616 J / v2f
Cross-multiplying and rearranging the equation gives us:
1.46616 J = 0.3672 kg * v2f
Dividing both sides by 0.3672 kg:
v2f = 1.46616 J / 0.3672 kg ≈ 3.997 m/s
Therefore, the speed of the second cart after impact is approximately 3.997 m/s.
(c) To find the speed of the two-cart center of mass, we can use the conservation of momentum equation:
(m1)(v1i) + (m2)(v2i) = (m1)(v1f) + (m2)(v2f)
Plugging in the known values, we get:
(0.34 kg)(1.5 m/s) + (m2)(0 m/s) = (0.34 kg)(0.42 m/s) + (m2)(3.997 m/s)
Simplifying the equation gives us:
0.51 kg = 0.1428 kg + (m2)(3.997 m/s)
Now, solve for m2 by subtracting 0.1428 kg from both sides:
0.3672 kg = (m2)(3.997 m/s)
Divide both sides by 3.997 m/s to solve for m2:
m2 = 0.3672 kg / 3.997 m/s ≈ 0.0919 kg
Therefore, the mass of the second cart is approximately 91.9 g.
The speed of the two-cart center of mass can be calculated by substituting the value of m2 into the equation:
0.51 kg = 0.1428 kg + (0.0919 kg)(v2f)
Simplifying the equation gives us:
0.3672 kg = 0.0919 kg * v2f
Solving for v2f by dividing both sides by 0.0919 kg:
v2f = 0.3672 kg / 0.0919 kg ≈ 3.998 m/s
Therefore, the speed of the two-cart center of mass is approximately 3.998 m/s.