Answer:
A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^
B) K_total = 373.08 × 10^(-15) J
Explanation:
We are given;
Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg
Mass of first particle; m1 = 5.03 × 10^(-27) kg
Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^
Mass of second particle; m2 = 8.47 × 10^(-27) kg
Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^
Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;
M = m1 + m2 + m3
1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3
m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))
m3 = 4.8 × 10^(-27) kg
A) Applying law of conservation of momentum, we have;
MV = (m1 × v1) + (m2 × v2) + (m3 × v3)
Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.
Thus, we now have;
0 = (m1 × v1) + (m2 × v2) + (m3 × v3)
We want to find the velocity of the third particle v3. Let's make it the subject of the formula;
v3 = [(m1 × v1) + (m2 × v2)]/(-m3)
Plugging in the relevant values, we have;
v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))
v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))
v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^
B) Formula for kinetic energy is;
K = ½mv²
Now,total kinetic energy is;
K_total = K1 + K2 + K3
K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²
K1 = 90.54 × 10^(-15) J
K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²
K2 = 67.76 × 10^(-15)
To find K3, let's first find the magnitude of v3 because it's still in vector form.
Thus;
v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]
v3 = 9.46 × 10^(6) m/s
K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²
K3 = 214.78 × 10^(-15) J
K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))
K_total = 373.08 × 10^(-15) J
the heat (energy) you must give to it is
is the specific heat of that substance (given in J/(g*Celsius))