Is my work correct?
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.
My ans.: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.)
To verify the correctness of your work, let's go through each step and calculation and compare it to the correct method.
a) Find the velocity of the 3.75kg mass before impact:
You correctly used the principle of conservation of energy to determine the velocity. The potential energy (Ep) of the 3.75kg mass will be equal to its kinetic energy (Ek). So, the equation you derived to find the velocity is correct:
Ep = mgh
Ek = (1/2)mv^2
Setting them equal:
(1/2)mv^2 = mgh
Now solving for v:
v^2 = 2gh
v = √(2gh)
Plugging in the values:
v = √(2 * 9.81 m/s^2 * 0.545m)
v ≈ 3.26 m/s
So, the correct velocity before impact is approximately 3.26 m/s, not 3.27 m/s as you stated.
b) Find the common velocity of the mass after impact:
To find the common velocity after impact, you correctly used the principle of conservation of kinetic energy. The kinetic energy lost by the first mass before impact is equal to the kinetic energy gained by the combined masses after impact. Your equation is correct:
(1/2)m1(v1)^2 = (1/2)(m1 + m2)v2^2
Plugging in the values:
(1/2)(3.75 kg)(3.26 m/s)^2 = (1/2)(3.75 kg + 2.4 kg)v2^2
19.94 = 6.15v2^2
Solving for v2:
v2^2 = 19.94 / 6.15
v2 ≈ 2.56 m/s
So, the correct common velocity after impact is approximately 2.56 m/s, not 2.55 m/s as you stated.
c) Find the loss of kinetic energy on impact:
To find the loss of kinetic energy on impact, you correctly used the formula for kinetic energy. However, there is a slight calculation error:
Ek2 = (1/2)(m1 + m2)v2^2
Plugging in the values:
Ek2 = (1/2)(3.75 kg + 2.4 kg)(2.56 m/s)^2
Ek2 = 3.075(2.56)^2
Calculating:
Ek2 ≈ 19.75 J
So, the correct loss of kinetic energy on impact is approximately 19.75 J, not 19.99 J as you stated.
d) Find the height through which the center of gravity of the system will rise:
To find the height through which the center of gravity of the system will rise, you correctly used the principle of conservation of potential energy. The potential energy before impact is equal to the potential energy after impact. Your equation is correct:
m1gh1 = (m1 + m2)gh2
Plugging in the values:
(3.75 kg)(9.81 m/s^2)(0.545 m) = (3.75 kg + 2.4 kg)(9.81 m/s^2)h2
20.05 = 6.15h2
Solving for h2:
h2 = 20.05 / 6.15
h2 ≈ 3.26 m
So, the correct height through which the center of gravity of the system will rise is approximately 3.26 m, not 0.332 m as you stated.
e) Find the tension on the string of the 3.75kg mass before impact:
To find the tension on the string, you correctly recognized that it is equal to the centrifugal force exerted by the 3.75kg mass. However, there is a slight calculation error:
Fc = (mv^2) / r
Plugging in the values:
Fc = (3.75 kg)(3.26 m/s)^2 / 2.1 m
Fc ≈ 19.32 N
So, the correct tension on the string of the 3.75kg mass before impact is approximately 19.32 N, not 19.09 N as you stated.
In summary, most of your steps and calculations are correct, but there are a few minor errors in the numerical values. It is important to double-check all calculations to ensure accuracy.