Respuesta :
Answer:
The speed of the package of mass m right before the collision [tex]= 7.668\ ms^-1[/tex]
Their common speed after the collision [tex]= 2.56\ ms^-1[/tex]
Height achieved by the package of mass m when it rebounds [tex]= 0.33\ m[/tex]
Explanation:
Have a look to the diagrams attached below.
a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.
[tex]K_{initial} + U_{initial} = K_{final}+U_{final}[/tex]
where [tex]K[/tex] is Kinetic energy and [tex]U[/tex] is Potential energy.
[tex]K= \frac{mv^2}{2}[/tex] and [tex]U= mgh[/tex]
Considering the fact [tex]K_{initial} = 0\ and U_{final} =0[/tex] we will plug out he values of the given terms.
So [tex]V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1[/tex]
Keypoints:
- Sum of energies and momentum are conserved in all collisions.
- Sum of KE and PE is also known as Mechanical energy.
- Only KE is conserved for elastic collision.
- for elastic collison we have [tex]e=1[/tex] that is co-efficient of restitution.
KE = Kinetic Energy and PE = Potential Energy
b.Now when the package stick together there momentum is conserved.
Using law of conservation of momentum.
[tex]m_1V_1(i) = (m_1+m_2)V_f[/tex] where [tex]V_1{i} =7.668\ ms^-1[/tex].
Plugging the values we have
[tex]m\times 7.668 = (3m)\times V_{f}[/tex]
Cancelling m from both sides and dividing 3 on both sides.
[tex]V_f = 2.56\ ms^-1[/tex]
Law of conservation of energy will be followed over here.
c.Now the collision is perfectly elastic [tex]e=1[/tex]
We have to find the value of [tex]V_{f}[/tex] for m mass.
As here [tex]V_{f}=-2.56\ ms^-1[/tex] we can use that if both are moving in right ward with [tex]2.56[/tex] then there is a [tex]-2.56[/tex] velocity when they have to move leftward.
The best option is to use the formulas given in third slide to calculate final velocity of object [tex]1[/tex].
So
[tex]V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1[/tex]
Now using law of conservation of energy.
[tex]K_{initial} + U_{initial} = K_{final}+U_{final}[/tex]
[tex]\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh[/tex]
[tex]\frac{v(f1)^2}{2g} = h[/tex]
[tex] h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m[/tex]
The linear momentum is conserved before and after this perfectly elastic collision.
So for part a we have the speed [tex]=7.668\ ms^-1[/tex] for part b we have their common speed [tex]=2.56\ ms^-1[/tex] and for part c we have the rebound height [tex]=0.33\ m[/tex].
(a) The initial speed of the package m, before collision, is 7.67 m/s.
(b) The final speed of the two packages after collision if they stick together is 2.55 m/s.
The mechanical energy of the system is not conserved and the change is -19.66m J.
(c) The height reached by the mass m after rebound is 3.0 m
For the perfectly elastic collision, the total momentum and kinetic energy of the system are conserved.
The given parameters;
- height, h = 3.0 m
- mass of the first package, = m
- mass of the second package, = 2m
The initial speed of the package m, before collision, is calculated as follows;
[tex]\frac{1}{2} mu_1^2 = mgh\\\\u_1^2 = 2gh\\\\u_1 = \sqrt{2gh} \\\\u_1 = \sqrt{2\times 9.8 \times 3} \\\\u_1 = 7.67 \ m/s[/tex]
The final speed of the two packages after collision if they stick together is determined by applying the principle of conservation of linear momentum as follows;
[tex]m_1 u_1 + m_2u_2 = v(m_1 + m_2 )\\\\m(7.67) + 2m(0) = v(m + 2m)\\\\7.67 m = 3mv\\\\v = \frac{7.67m}{3m} \\\\v = 2.55 \ m/s[/tex]
The change in mechanical energy before and after collision is calculated as;
[tex]\Delta K.E = \frac{1}{2}(m_1 + m_2)v^2 - [\frac{1}{2} m_1u_1^2 - \frac{1}{2} m_2u_2^2]\\\\\Delta K.E = \frac{1}{2}(m + 2m)(2.55^2) - (\frac{1}{2} m\times 7.67^2)- (0)\\\\\Delta K.E = -19.66 m \ J[/tex]
Thus, the mechanical energy of the system is not conserved.
The final velocity of the package m when the collision is perfectly elastic is calculated as;
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\m(7.67) + 2m(0) = mv_1 \ + \ 2mv_2\\\\7.67m = mv_1 + 2mv_2[/tex]
Apply one dimensional velocity formula;
[tex]u_1 + u_2 = v_1 + v_2\\\\7.67 + 0 = v_1 + v_2\\\\ v_2 = 76.7 - v_1[/tex]
substitute the value of [tex]v_2[/tex] in the first equation;
[tex]7.67 m = mv_1 + 2m(7.67 - v_1)\\\\7.67 m = mv_1 +15.34m - 2mv_1\\\\7.67m- 15.34m = m(v_1 - 2v_1)\\\\-7.67 m = m(-v_1)\\\\v_1 = 7.67 \ m/s[/tex]
The height reached by the mass m at the calculated rebound speed is calculated as;
[tex]v^2 = v_0^2 + 2gh\\\\v^2 = 2gh\\\\h = \frac{v^2}{2g} \\\\h = \frac{(7.67)^2}{2 \times 9.8} \\\\h = 3 \ m[/tex]
Thus, for the perfectly elastic collision, the total momentum and kinetic energy of the system are conserved.
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