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.

Question

a. Find the speed of the thorium nucleus. v=

b. Find the direction of the motion of the thorium nucleus.

Answer 1

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)

Answer 2

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

Answer 3

Np Bro

Answer 4

english pls