HW 7 - Rotation Begin Date: 6/30/2018 12:01:00 AM - Due Date: 7/11/2018 11:59:00

Question

HW 7 - Rotation Begin Date: 6/30/2018 12:01:00 AM - Due Date: 7/11/2018 11:59:00 PM End Date: 7/11/2018 11:59:00 PM (4%) Problem 17: A bullet is fired horizontally into an initially stationary block of wood suspended by a string and remains embedded in the block. The bullet's mass is m = 0.0085 kg, while that of the block is M = 0.92 kg. After the collision the block/bullet system swings and reaches a maximum height of h- 0.95 m above its initial height. Neglect air resistance × 14% Part (a) Enter an expression for the speed of the block/bullet system immediately after the collision in terms of defined quantities and g. 14 % Part (b) Find the speed of the block/bullet system, in meters per second, immediately after the collision 14% Part (c) Enter an expression for the initial speed of the bullet in terms of defined quantities and g 14% Part (d) Find the initial speed of the bullet, in meters per second 14% Part (e) Find the initial kinetic energy of the bullet, in joules 14% Part (f) Enter an expression for the kinetic energy of the block/bullet system immediately after the collision in terms of defined quantities and g > ? 14% Part (g) Calculate the ratio, expressed as a percent of the kinetic energy of the block bullet system immediately after the collision to the initial kinetic energy of the bullet. Grade Summa Deductions Potential ry 0% 100% cos) atanO)acotanOsinh) Degrees O Radians Submissions Attempts remaining: 10 (0% per attempt) detailed view tan() | ?| ( acos) sin cotan) asin) 1 23 0 coshO tanh() cotanhO END BACKSPACE CLEAR Submit Hint I give up!

Explanation / Answer

A)

let the speed of the combination after collision be "V"

h = height gained

(m + M) = total mass of combination

using conservation of energy

Potential energy gained = kinetic energy lost

(m + M) g h = (0.5) (m + M) V2

V = sqrt(2gh)

G)

V = sqrt(2gh) = sqrt(2 x 9.8 x 0.95) = 4.32 m/s

v = speed of bullet before collision

using conservation of momentum

mv = (m + M) V

0.0085 v = (0.0085 + 0.92) (4.32)

v = 472 m/s

KEi = kinetic energy before collision = (0.5) m v2 = (0.5) (0.0085) (472)2 = 946.83 J

KEf = kinetic energy after collision = (0.5) (m + M) V2 = (0.5) (0.0085 + 0.92) (4.32)2 = 8.66 J

ratio = KEf x 100/KEi = 8.66 x 100/946.83 = 0.915 %

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