Asked by Peter
                A 238U nucleus is moving in the x-direction at 4.8×105m/s when it decays into an alpha particle (4He) and a 234Th nucleus. The alpha moves at 1.2×107m/s at 36∘ above the x-axis.
a. Find the speed of the thorium nucleus. v=
b. Find the direction of the motion of the thorium nucleus.
            
        a. Find the speed of the thorium nucleus. v=
b. Find the direction of the motion of the thorium nucleus.
Answers
                    Answered by
            Damon
            
    original x momentum = 238 * 4.8*10^5
original y momentum = 0
final y momentum is still zero
0 = 234 Vt + 4 Vhe
final x momentum is still
238 * 4.8*10^5
= 234 Ut + 4 Uhe
but we know
Uhe = 1.2 *10^7 cos 36 = you do it
Vhe = 1.2 *10^7 sin 36 = you do it
Then go back and solve for Vt and Ut
speed = sqrt(Ut^2+Vt^2)
tan angle = (Vt/Ut)
    
original y momentum = 0
final y momentum is still zero
0 = 234 Vt + 4 Vhe
final x momentum is still
238 * 4.8*10^5
= 234 Ut + 4 Uhe
but we know
Uhe = 1.2 *10^7 cos 36 = you do it
Vhe = 1.2 *10^7 sin 36 = you do it
Then go back and solve for Vt and Ut
speed = sqrt(Ut^2+Vt^2)
tan angle = (Vt/Ut)
                    Answered by
            Peter
            
    your process is wrong I ended up with the wrong answer. of 2.9*10^5 instead of THE REAL ANSWER 3.4*10^5 
and 21 degrees.
thanks
    
and 21 degrees.
thanks
                    Answered by
            Peter
            
    Np Bro
    
                    Answered by
            spongbob
            
    english pls
    
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