Question

In the figure below, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle ?1 = 66.0

Answer 1

Let m1, m2 be the masses of particles 1 and 2, v1 and v2 their velocities after the collision, u the velocity of particle 1 before collision, particle 2 being initially at rest.
By the law of conservation of momentum, we can write:
vector (m1.u) = vector (m1.v1) + vector (m2.v2)
Let's represent graphically the addition of two vectors m1v1, m2v2 by drawing a parallelogram OPRQ in which
vector OP = vector m1v1
vector OQ = vector m2v2
vector OR = vector m1u
We get:
...angle ROP = ?1 = 66.0°
...angle ROQ = ?2 = 55.0°
...angle ORQ = angle ROP = 66.0°
...angle OQR = 180° - (66.0° + 55.0°) = 59.0°, and
...RQ = OP
Applying the sines law to triangle OQR:
OQ/sin66° = RQ/sin55° = OR/sin59°
OQ/sin66° = OP/sin55° = OR/sin59°
m2v2/sin66° = m1v1/sin55° = m1u/sin59°
(a)
m2v2/sin66° = m1v1/sin55°
v1 = v2(m2/m1)(sin55°/sin66°)
v1 = 1.30×10^5×(16/4)×(sin55°/sin66°)
v1 = 5.6921×10^5×sin55° = 4.66×10^5 m/s
The final speed of the alpha particle is 4.66×10^5 m/s
(b)
m2v2/sin66° = m1u/sin59°
u = v2(m2/m1)(sin59°/sin66°)
u = 1.30×10^5×(16/4)×(sin59°/sin66°)
u = 5.6921×10^5×sin59° = 4.88×10^5 m/s
The initial speed of the alpha particle is 4.88×10^5 m/s

Answer 2

a.4.1629*10^(5) m/s
b.5.2396*10^5 m/s

Answer 3

Let m1, m2 be the masses of particles 1 and 2, v1 and v2 their velocities after the collision, u the velocity of particle 1 before collision, particle 2 being initially at rest.
By the law of conservation of momentum, we can write:
vector (m1.u) = vector (m1.v1) + vector (m2.v2)
Let's represent graphically the addition of two vectors m1v1, m2v2 by drawing a parallelogram OPRQ in which
vector OP = vector m1v1
vector OQ = vector m2v2
vector OR = vector m1u
We get:
...angle ROP = ?1 = 66.0°
...angle ROQ = ?2 = 55.0°
...angle ORQ = angle ROP = 66.0°
...angle OQR = 180° - (66.0° + 55.0°) = 59.0°, and
...RQ = OP
Applying the sines law to triangle OQR:
OQ/sin66° = RQ/sin55° = OR/sin59°
OQ/sin66° = OP/sin55° = OR/sin59°
m2v2/sin66° = m1v1/sin55° = m1u/sin59°
(a)
m2v2/sin66° = m1v1/sin55°
v1 = v2(m2/m1)(sin55°/sin66°)
v1 = 1.30×10^5×(16/4)×(sin55°/sin66°)
v1 = 5.6921×10^5×sin55° = 4.66×10^5 m/s
The final speed of the alpha particle is 4.66×10^5 m/s
(b)
m2v2/sin66° = m1u/sin59°
u = v2(m2/m1)(sin59°/sin66°)
u = 1.30×10^5×(16/4)×(sin59°/sin66°)
u = 5.6921×10^5×sin59° = 4.88×10^5 m/s
The initial speed of the alpha particle is 4.88×10^5 m/s