Two bodies hang on a string 2.1m long. The two bodies have a mass of 3.75kg and 2.4kg respectively. If the 3.75kg mass is raised to a height of 0.545m and released, and if upon impact the mass adhere.
Mass1 = 3.75 kg
Mass2 = 2.4 kg
String length = 2.1 m
M1 raised = 0.545m
a) Find the velocity of the 3.75kg mass before impact.
Because the potential energy (Ep) of the 3.75kg mass will be equal to its kinetic energy (Ek), the following equation can be derived to find the velocity:
E_k=1/2 mv^2
E_p=mgh
1/2 mv^2=mgh
v^2=2gh
v=¡Ì2gh
v=¡Ì(2X9.81 m⁄s^2 X0.545m)
v=¡Ì(10.69 m^2⁄s^2 )
Velocity before impact=3.27 m⁄s (Ans.)
b) Find the common velocity of the mass after impact.
Lost inE_(k1 ) before impact=Gain inE_k2 after impact
1/2 M_1 (v_1 )^2=1/2 (M_1+M_2 )X(¡¼v_2)¡½^2
0.5X3.75kgX(3.27 m⁄s)^2=0.5X(3.75kg+2.4kg)X¡¼v_2¡½^2
20.049=3.075X¡¼v_2¡½^2
¡¼v_2¡½^2=2.049/3.075
¡Ì(¡¼v_2¡½^2 )=¡Ì6.52
Velocity after impact=2.55 m⁄s (Ans.)
c) Find the lost of kinetic energy on impact.
Loss of E_k1 on impact=¡¼Ek¡½_2 after impact
E_k2=1/2(m_1+m_2)¡¼v_2¡½^2
E_k2=0.5X(3.75kg+2.4kg)X(2.55 m⁄s)^2
E_k2=3.075X¡¼2.55¡½^2
Kinetic Energy after impact=19.99J (Ans.)
d) Find the height through which the center of gravity of the system will rise.
E_p1 before impact=E_p2 after impact
m_1 gh_1=(m_1+m_2 )gh_2
3.75kgX9.81X0.545m=(3.75kg+2.4kg)X9.81Xh_2
20.05=60.33Xh_2
h_2=20.05/60.33
Final height the system will rise=0.332m (Ans.)
e) Find the tension on the string of the 3.75kg mass before impact.
Tension on the string is equal to the centrifugal force exerted by the 3.75kg mass.
F_c=(mv^2)/r
F_c=(3.75X¡¼3.27¡½^2)/2.1
F_c=40.098/2.1
F_c=19.09N (Ans.)
You do not seem to have completed the question. By the way, if the two bodies are hanging on the same string, how can ther be an im[act?
Perhaps they are hanging from the same hook 2.1 meters below the hook but on different strings?
Then the 3.75 kg one is lifted up .545 m
this gains U = m g h = 3.75*9.8*.545
U = 20 Joules
let it go
then just before they hit
(1/2) m v^2 = 20
v^2 = 40/3.75 = 10.7
v = 3.27 m/s
momentum = m v = 3.75*3.27 = 12.2 kg m/s
same momentum after collision but mass = 3.75+2.4 = 6.15
so v after crash = 12.2/6.15 = 2 m/s
(1/2)mv^2 = (1/2)(6.15)(4) = 12.3 Joules
If you want to find out high high they go stuck together
6.15 * 9.8 * h = 12.3
h = .204 meters
You figured it out, thanks Damon. But how come my work didn't get posted with the question... I did it like you up until the 12.2kg m/s part... Now i know where I went wrong.. T-Y
Henri
This was the original post...
Two bodies hang on a string 2.1m long. The two bodies have a mass of 3.75kg and 2.4kg respectively. If the 3.75kg mass is raised to a height of 0.545m and released, and if upon impact the mass adhere.
Mass1 = 3.75 kg
Mass2 = 2.4 kg
String length = 2.1 m
M1 raised = 0.545m
a) Find the velocity of the 3.75kg mass before impact.
Because the potential energy (Ep) of the 3.75kg mass will be equal to its kinetic energy (Ek), the following equation can be derived to find the velocity:
E_k=1/2 mv^2
E_p=mgh
1/2 mv^2=mgh
v^2=2gh
v=¡Ì2gh
v=¡Ì(2X9.81 m⁄s^2 X0.545m)
v=¡Ì(10.69 m^2⁄s^2 )
Velocity before impact=3.27 m⁄s (Ans.)
b) Find the common velocity of the mass after impact.
Lost inE_(k1 ) before impact=Gain inE_k2 after impact
1/2 M_1 (v_1 )^2=1/2 (M_1+M_2 )X(¡¼v_2)¡½^2
0.5X3.75kgX(3.27 m⁄s)^2=0.5X(3.75kg+2.4kg)X¡¼v_2¡½^2
20.049=3.075X¡¼v_2¡½^2
¡¼v_2¡½^2=2.049/3.075
¡Ì(¡¼v_2¡½^2 )=¡Ì6.52
Velocity after impact=2.55 m⁄s (Ans.)
c) Find the lost of kinetic energy on impact.
Loss of E_k1 on impact=¡¼Ek¡½_2 after impact
E_k2=1/2(m_1+m_2)¡¼v_2¡½^2
E_k2=0.5X(3.75kg+2.4kg)X(2.55 m⁄s)^2
E_k2=3.075X¡¼2.55¡½^2
Kinetic Energy after impact=19.99J (Ans.)
d) Find the height through which the center of gravity of the system will rise.
E_p1 before impact=E_p2 after impact
m_1 gh_1=(m_1+m_2 )gh_2
3.75kgX9.81X0.545m=(3.75kg+2.4kg)X9.81Xh_2
20.05=60.33Xh_2
h_2=20.05/60.33
Final height the system will rise=0.332m (Ans.)
e) Find the tension on the string of the 3.75kg mass before impact.
Tension on the string is equal to the centrifugal force exerted by the 3.75kg mass.
F_c=(mv^2)/r
F_c=(3.75X¡¼3.27¡½^2)/2.1
F_c=40.098/2.1
F_c=19.09N (Ans.)
Copy and paste is not allowed on this site. That is probably the problem.
To solve this problem, we need to apply principles of conservation of energy and momentum.
a) The velocity of the 3.75kg mass before impact can be found using the equation v = √(2gh), where g is the acceleration due to gravity and h is the height. Plugging in the values, we get v = √(2 * 9.81 m/s² * 0.545m) = √(10.69 m²/s²) = 3.27 m/s.
b) The common velocity of the mass after impact can be found using the principle of conservation of momentum. The momentum before impact is given by M1 * v1, where M1 is the mass of the 3.75kg mass and v1 is its velocity before impact. The total momentum after impact is (M1 + M2) * v2, where M2 is the mass of the 2.4kg mass and v2 is the common velocity after impact. Since the masses adhere, their velocities are the same. Setting the initial and final momenta equal, we have M1 * v1 = (M1 + M2) * v2. Plugging in the values, we get 3.75kg * 3.27 m/s = (3.75kg + 2.4kg) * v2. Solving for v2, we find v2 = 2.55 m/s.
c) The lost kinetic energy on impact can be found by comparing the initial and final kinetic energies. The initial kinetic energy of the system is given by 1/2 * M1 * v1^2, and the final kinetic energy is given by 1/2 * (M1 + M2) * v2^2. Plugging in the values, we get 1/2 * 3.75kg * (3.27 m/s)^2 = 1/2 * (3.75kg + 2.4kg) * (2.55 m/s)^2. Simplifying the equation, we find that the kinetic energy after impact is 19.99J.
d) The height through which the center of gravity of the system will rise can be found using the principle of conservation of potential energy. The initial potential energy is given by M1 * g * h1, where h1 is the initial height. The final potential energy is given by (M1 + M2) * g * h2, where h2 is the final height. Setting the initial and final potential energies equal, we have M1 * g * h1 = (M1 + M2) * g * h2. Plugging in the values, we get 3.75kg * 9.81 m/s^2 * 0.545m = (3.75kg + 2.4kg) * 9.81 m/s^2 * h2. Solving for h2, we find h2 = 0.332m.
e) The tension on the string of the 3.75kg mass before impact can be found by considering the centrifugal force exerted by the mass. The centrifugal force is given by Fc = (m * v^2) / r, where m is the mass and v is the velocity. Plugging in the values, we get Fc = (3.75kg * (3.27 m/s)^2) / 2.1m. Simplifying the equation, we find that the tension on the string is 19.09N.