Question

proton, moving with a velocity of St, col ides elastically with another proton that is initiall y est. If th e two protons have equal speeds a fter the collision. Q7:- Fird the speed of each proton a ter the collision (A)0.71ms-1 (B) 1.4ms-1 (C) 2.11ns" (D) 2.8ms-1 (E) 3.5ms-1 Q8:- The direction of the velocity vectors after the collision. (A)S5-35) (B)(30.-60) (C(45-45) (D)(G0,-30) (E)(50;-40) . of mass-,5kg,m2.2 kg, and m3-3 kg are located in xy plane as 7 Q9:- T hree particle 3.1)m. (-2,1)m, and (3.-1)m, respectively. Find the coordinate of the center of mass. (A)(2, 1.1)m (B)(2,0.8)m (c, 1)m D)(2.0.4)m (E2,0.6)m 10 A5 ke particle has a velocity of 31 + 4 (m/S), Find the magnitude of its total Q10:- momertum (A)15kgnus-1 (E) 20kgms-1Dskgrn.5-1 (D)5kgms-1 (E) 10kgrns-1 Q11:-Two vectors are given by A+ and B-3t+1+R.. Find the angle between A and (A)19.3 (B)31.5 (C)82.3(D)72.5 (E)64.8 Q12:- A worker is to paint the walls ()of a sq are room 7.00 ft high an d 13.0 ft along each side. What surface area in square meters must she cover?1m 3.28ft (A)33.8m2 (B)35.7m2 (C)29.7m2 (D)36.82 (E)32.7m2 Q 13:- A ball is thrown vertically upward from the gro und with an initial spee d of 15 m/s. Determine the velocity of the ball att- is. (A) 1ms (B)2ms1 (C)3ms1 (D) 4ms1 (E) 5ms-1 Q14:-At what angle to the stadium (st) ground a footbal I is kicked (S). If it reached a rnaximum height of (10 m), and a maximum horizontal dis tance of ( 60 )? (4)45 (B) 26.6(C) 531 (D) 33.7 (E) 38.6 015:- A 9.00-kg hanging weight is connected by a string over a pulley to a 5.00-kg block that is sliding on a la table Fig. If the coefficient of kinetic friction is 0.200, finc the tension in the string. (A)39.2N (D)34.55N (E)35. 56N 5.00 kg (B)38 57N (C)37.23N 9 00kg

Answer 1

Nomenclature is like this

u1xi = initial speed of moving proton in x-direction = 5 m/sec

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 = 5 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
5 - 0 = V*cos $\theta$1 + V*cos $\theta$2
V*cos $\theta$1 + V*cos $\theta$2 = 5

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 = 5

V*cos $\theta$ + V*cos $\theta$ = 5

2*V*cos $\theta$ = 5

V = 5/(2*cos 45 deg)

V = 3.5 m/sec

Correct option of Q7 is option E

Q9.

A.

X-coordinate of Center of mass is given by:

x = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3)

x = (5*3 + 2*(-2) + 3*3)/(5 + 2 + 3)

x = 2 m

B.

Y-coordinate of Center of mass is given by:

y = (m1y1 + m2y2 + m3y3)/(m1 + m2 + m3)

y = (5*1 + 2*1 + 3*(-1))/(5 + 2 + 3)

y = 0.4 m

Center of mass = (x, y) = (2, 0.4) m

Correct option is D.

Q10.

momentum is given by:

P = m*V

Given that m = 5 kg

V = 3 i + 4 j

So,

P = m*V = 5*(3 i + 4 j)

P = 15 i + 20 j

Magnitude of momentum will be:

|P| = sqrt (15^2 + 20^2) = 25 kg-m/sec

Correct option is C.

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