Question

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)
A. What is the magnitude of the change in the momentum, Δp1, of mass M1?
B. What is the change in the total momentum of the pair?
C. What is the magnitude of the change in the momentum Δp2, of mass M2?

Answer 1

Answer:

a). ΔP1=-2.4 $x10^{3} \frac{D*m}{s}$

b). Pp=0 F=0

c). ΔP2=2.4 $x10^{3} \frac{D*m}{s}$

Explanation:

Initial momentum

$P_{1}=m_{1}*v_{i1}$

Final momentum

$P_{1f}=m_{1}*v_{f1}=-m_{1}*v_{i1}$

The change of momentum m1 is:

a).

ΔP1=$P_{1f}-P_{1}$

ΔP1=$-m_{1}*v_{i1}-m_{1}*v_{i1}$

ΔP1=$-2*m_{1}*v_{i1}$

ΔP1=$-2*6 D*200\frac{m}{s}$

ΔP1=$-2.4x10^{3}\frac{D*m}{s}$

b).

The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero

P=0

Fx=0

c).

ΔP1+ΔP2=0

ΔP2=-ΔP1

ΔP2=-$-2.4x10^{3}\frac{D*m}{s}$

ΔP2=$2.4x10^{3}\frac{D*m}{s}$