Question

A proton, moving with a velocity of 21, collides elastically with another proton that is initially rest. If the two protons have equal speeds, after the collision. Q7:- Find the speed of each proton after the collision (A)0.71ms-1 (B) 1.4ms (C) 2.1ms1 (D) 2.8ms1(E) 3.5ms-1 Q8:-The direction of the velocity vectors after the collision. (A)(SS,-35) (B)(30,-60) (C)(45',-45) (D)(60,-30) (E) (50',-40 )

Answer 1

Nomenclature is like this

u1xi = initial speed of moving proton in x-direction = 2

u2xi = initial speed of stationary proton in x-direction = 0

u1yi = initial speed of moving proton in y-direction = 0

u2yi = initial speed of stationary proton in y-direction = 0

V1, V2 = final speed of both protons = V

$\theta$1 = final direction of initially moving proton

$\theta$2 = final direction of initially stationary proton

Also see that one proton will go Counterclockwise from +ve axis and other proton will go clockwise from +x-axis, which means  $\theta$1 will be +ve and $\theta$2 will be -ve.

In a perfectly elastic collision

Using momentum conservation in x direction,
Pi = Pf
m1*u1xi + m2*u2xi = m1*V1x + m2*V2x
given that m1 = m & m2 = m

u1xi = 2 m/sec & u2xi = 0 (since stationary)

V1x = V*cos $\theta$1 & V2x = V*cos $\theta$2
m*u1xi + m*u2xi = m*V*cos $\theta$1 + m*V*cos $\theta$2
2 - 0 = V*cos $\theta$1 + V*cos $\theta$2
V*cos $\theta$1 + V*cos $\theta$2 = 2

Using momentum conservation in y direction,
Pi = Pf
m1*u1yi + m2*u2yi = m1*V1y + m2*V2y
given that m1 = m & m2 = m

u1yi = 0 m/sec & u2yi = 0 m/sec

V1y = V*sin $\theta$1 & V2y = -V*sin $\theta$2
m*u1i + m*u2i = m*V*sin $\theta$1 - m*V*sin $\theta$2

0 - 0 = V*sin $\theta$1 - V*sin $\theta$2

V*sin $\theta$1 - V*sin $\theta$2 = 0

V*sin $\theta$1 = V*sin $\theta$2

sin $\theta$1 = sin $\theta$2

$\theta$1 = $\theta$2 = $\theta$

from given option we can see that

$\theta$ = 45deg

($\theta$1, $\theta$2) = (45, -45)

Correct answer of Q8. is option C.

Now Using above relation in 1st equation

V*cos $\theta$1 + V*cos $\theta$2 = 2

V*cos $\theta$ + V*cos $\theta$ = 2

2*V*cos $\theta$ = 2

V = 2/(2*cos 45 deg)

V = 1.4 m/sec

Correct option of Q7 is option B

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