1. If a hydrogen nucleus collides with velocity v, with a nucleus of Li' the latter splits into two...

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1. If a hydrogen nucleus collides with velocity v, with a nucleus of Li' the latter splits into two a particles (mass = 4 in ). Assume that the a 's fly off symmetrically with respect to the line of collision, with equal velocities, and calculate the angle 249 between them. Recall that in addition to the kinetic energy Er, of the proton, some amount of rest mass energy E , E>> EP , is liberated from the fissioning of the lithium nucleus and is imparted to the a particles. If E =14 MeV, and E,=0.2 MeV, what are the values of vc, and 2 qo ?  2. For the following system, if M is acted on by a force 13.= c cos (w t) , design the system such that M stays at rest throughout the motion. What conditions must be satisfied by m & k? The system only moves in the x coordinate (along the surface) without friction XXXXXXXXXXA particle moves in an attractive central force field such that F ( r)=-7 . If the particle is given a velocity u0 at a distance R away from the force center and in a direction a , find the dependence of u0 on a if u0 is the minimum velocity for escape.  / \ •-• "" XXXXXXXXXXFor an attractive central force given by F(r)=-7 , show that u( 0) , where u =1/r satisfies  (M 11" XXXXXXXXXX2C =0 , and 1 is the angular momentum. Plot graphs of the orbit for the following numerical values:  1— mC =3,0.15,0; and 12  MC 12  1=0.15 .

Robert · Accepted Answer

Mechanics Solutions
1. Given the problem as shown in following diagram.
φ
φ
α(m=4)
vp, Ep
vα
H+(m=1)
LI7(m=7)
α(m=4)
vα
Figure 1: Schematic for Problem 1
Initial Kinetic Energy KE=Energy of Proton=Ep = 0.2MeV .
Final KE=KE of 2 alpha particles=2
(
1
2 (4mp)v
2
α
)
By momentum balance, Initial Momentum=Final Momentum⇒ mpvpî = 2(4mp × (vαcosφî))
⇒ vp = 8vαcosφ
Also, from energy balance, Final KE=Initial KE+Fission Energy⇒ 2
(
1
2 (4mp)v
2
α
)
= (14 + 0.2)× (1.60217646× 10−13)J
Also, we have mp = 1.67262158× 10−27Kg. Using these, solving the three above equations for three unknowns, we get as
follows.
vp = ±6.19× 106m/s, vα = ±1.844× 107m/s, φ = 1.5289 rad.
OR
vp = ∓6.19× 106m/s, vα = ±1.844× 107m/s, φ = 1.613 rad.
However, we can neglect all the solution sets with negative velocities as the particles should always have velocities towards
right. So, we have final solution as
vp = 6.19× 106m/s, vα = 1.844× 107m/s, φ = 1.5289 rad.
2. Given the figure below and that M remains fixed. Also, force input for M is Px = ccosωt Drawing the Free Body Diagram,
we have as follows.
∑
F = ma⇒ mẍm = k(xM − xm) & MẍM = ccosωt−KxM − k(xM − xm)
Further,

1. If a hydrogen nucleus collides with velocity v, with a nucleus of Li' the latter splits into two a particles (mass = 4 in ). Assume that the a 's fly off symmetrically with respect to the line of...

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