The mass of a uranium-238 atom is 3.95*10^-25kg. A stationary uranium atom emits an alpha particle with a mass of 6.64*10^-27kg. If the alpha particle has a velocity of 1.42*10^4m/s, what is the recoil velocity of the uranium atom.

Ans: i tried MaV'a = -MbV'b
i happen to get 2.38*10^2m/s.....
is this the correct answer because the book said it is 2.48...

momentum is zero before and after

0 = mass U * speed U + mass alpha * speed alpha

To determine the recoil velocity of the uranium atom, we can apply the principle of conservation of momentum. This principle states that the total momentum before a collision is equal to the total momentum after the collision.

Let's denote the mass of the uranium atom as Ma, the velocity before emission as Va, the mass of the alpha particle as Mb, the velocity of the alpha particle as Vb, and the recoil velocity of the uranium atom as Va'.

According to conservation of momentum:
Ma * Va = Ma * Va' + Mb * Vb

We know the following values:
Ma = 3.95 * 10^-25 kg (mass of uranium atom)
Va = 0 m/s (initial velocity of uranium atom)
Mb = 6.64 * 10^-27 kg (mass of alpha particle)
Vb = 1.42 * 10^4 m/s (velocity of alpha particle)

Plugging in these values, the equation becomes:
3.95 * 10^-25 kg * 0 m/s = (3.95 * 10^-25 kg) * Va' + (6.64 * 10^-27 kg) * (1.42 * 10^4 m/s)

Simplifying this equation:
0 = (3.95 * 10^-25 kg) * Va' + (6.64 * 10^-27 kg) * (1.42 * 10^4 m/s)

Now, let's solve for Va':
Va' = -((6.64 * 10^-27 kg) * (1.42 * 10^4 m/s))/(3.95 * 10^-25 kg)
Va' = - (9.4448 * 10^-23 kg·m/s) / (3.95 * 10^-25 kg)

Calculating the value:
Va' = - 2.3933 * 10^2 m/s (approximately)

Hence, the recoil velocity of the uranium atom is approximately -2.3933 * 10^2 m/s. Note that the negative sign indicates that the uranium atom moves in the opposite direction of the emitted alpha particle.

The difference between your answer (2.38 * 10^2 m/s) and the book's answer (2.48 * 10^2 m/s) can be due to rounding differences or slight variations in calculations. Both results are reasonably close and within the same order of magnitude, so either value could be considered correct depending on the level of precision required.