Respuesta :
Answer:
(a) the momentum change is the same in magnitude but in opposite direction
(b) the kinetic energy change of the lighter fragment is 25 times as great as the kinetic energy change of the heavier fragment.
Explanation:
since there is no force applied to the shell , when it explodes
F= d(mv)/dt =0 therefore mv=constant
where F= force , m=mass , v=velocity , t=time
since the momentum is conserved , and denoting the heavier element as H and lighter as L
(mH+mL)* v initial = mH*vH+ mL*vL
since the system is initially at rest v initial = 0, then
mH*vH+ mL*vL = 0 → mL*vL = -mH*vH = M
therefore the momentum change if the lighter fragment is the same as the heavy one (a) in magnitude (but not in direction) , not (e) or (c)
since
- vL/vH= mH/mL = 25
the ratio of kinetic energy
KH/KL = 1/2 mH*vH² / (1/2 mL*vL)² = (mH/mL)*(vH/vL)² = 25* (1/25)² = 1/25
therefore
KL = 25 KH
thus (b) is also correct and not (d)
The momentum change of the lighter fragment is 25 times as great as the momentum change of the heavier fragment.
So. the correct option is (C).
What is momentum?
Momentum is defined as the product of mass to its velocity.
Solution :
Given,
[tex]m_1 = mass\; of \;particle 1\\m_2 = mass\; of\; particle 2[/tex]
[tex]P_h = m\\P_l= 25m[/tex]
Since the velocity is given zero.
And the mass of a particle [tex]P_l[/tex]25 times greater than the particle [tex]P_h[/tex]
[tex]P_h = 25 \times P_l[/tex]
Hence, the momentum change of the lighter fragment is 25 times as great as the momentum change of the heavier fragment will be the correct option.
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