0.764 v f. 0.280 kg 6.49 m/s 14.1 m 26 o 5. (1) Particle one of mass m initially

Question

0.764 v

f.

0.280 kg

6.49 m/s

14.1 m

26o

5. (1) Particle one of mass m initially moves to the right at v. Particle two of mass 2m is initially at rest. After they undergo an elastic collision, particle one moves directly down. What is the final speed of particle two? a. 0.250 v b. 0.577 v c. 0.500 v d.

0.764 v

6. (1) A 0.280-kg ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with one fourth the original speed of the first ball. The mass of the second ball is a 1.96 kg. b 0.560 kg. c 2.24 kg. d 0.840 kg. e 1.12 kg.

f.

0.280 kg

7. (1) A 60-kg man is standing on a 200-kg cart moving to the right at a speed of 5 m/s on frictionless tracks. If the man starts to walk to the left with a speed of 1 m/s relative to the cart, what is the final speed of the cart? a 3.18 m/s b 4.67 m/s c 5.23 m/s d

6.49 m/s

8. (1) A golf ball bounces to a height of 7.60 m after its second bounce after being dropped onto the sidewalk from a 20.5-m high pedestrian overpass. How high does the golf ball bounce after its first collision with the sidewalk? a 10.9 m b 12.5 m c 13.8 m d d15.6 m e

14.1 m

9. (1) Particle one of mass m initially moves to the right at v. Particle two of mass 3m is initially at rest. After they undergo an elastic collision, particle one moves directly down. In what direction does particle two move relative to the initial direction of particle one? a 30o b 60o c 63o d 35o e

26o

10. (1) A cannon fires a 2-kg exploding projectile across a level field on a windless day. The projectile leaves the cannon at a speed of 30 m/s at an angle of 60 degrees with respect to the horizontal. When the projectile reaches its maximum height, it explodes into two 1-kg pieces. One of the pieces falls straight down. How far away from the cannon does the second piece land? a 34 m b 40 m c 79 m d 92 m e 120 m

Explanation / Answer

In an elastic collision, both energy and momentum are conserved.

So, total initial momentum = total final momentum.

velocity of 1 before collision V i m/s   [i represents x-coordinate]

Let velocity of particle 1 after collision = -X j m/s      [since it is given that the particle is going down, it's velocity has only negative y-coordinate].

Let velocity of particle 2 after collision = Y i + Z j m/s

Initial momentum = m* (V i) + 2m*(0) = mV i

Final momentum = m*(-X) j + 2m*( Y i + Z j)

= 2mY i + m(2Z-X) j m/s

Equating i and j components of initial and final momentums,

2Y = V

Y = V/2

2Z-X = 0

Z = X/2

Also, initial energy = final energy.

Magnitude of final velocity of 2m = sqrt(Y^2 + Z^2)

Initial energy = 1/2 * m*V^2 + 1/2 * 2m*0

= 1/2 * mv^2

Final energy = 1/2 * m*(X^2) + 1/2 * 2m*(sqrt(Y^2 + Z^2))^2

= 1/2 * m * (X^2 + 2Y^2 + 2Z^2)

substituting Y = V/2 and Z = X/2

final energy = 1/2 * m* (X^2 + 2*V^2 /4 + 2*X^2 /4)

= 1/2 * m * (3X^2 /2 + V^2 /2)

Initial energy = final energy.

V^2 = 3X^2 /2 + V^2 /2

V^2 /2 = 3X^2 /2

X = V/sqrt(3) = 0.577 V

Magnitude of final velocity = sqrt((V/2)^2 + (V/2*sqrt(3))^2)

= sqrt(V^2 /4 + V^2 /12)

= sqrt(V^2 /3)

= V * 0.577

Note: In chegg more than single questions can't be asked as one question.

Please post each of the remaining questions separately as individual questions. Thank you

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