Question

A 10.0 gram bullet collides with a motionless 1.00 kilogram block of wood and becomes imbedded in the block. The block is suspended from above by strings which are 1.00 meter long (ignore the mass of the strings). After the collision the block swings to a point where the angle formed by the strings is 25.0 degrees to the vertical. a) what is the potential of the block/bullet combination relative to its original position? b) what is the kinetic energy of the block/bullet combination just after the collision? c) what is the velocity of the block/bullet combination just after the collision? d) what is the velocity of the bullet just before the collision? e) how much work did the bullet do on the block?

Answer 1

a)

L = length of the string = 1 m

= angle moved = 25 deg

h = height gained above the original position = L (1 - Cos)

h = 1 (1 - Cos25)

h = 0.094 m

Potential energy gained is given as

PE = (M + m) gh

PE = (1 + 0.010) (9.8) (0.094)

PE = 0.93 J

b)

Using conservation of energy

kinetic Energy of the combination just after collision = Potential energy gained

KE = 0.93 J

c)

Using conservation of energy

kinetic Energy of the combination just after collision = Potential energy gained

(0.5) (m + M) V² = (m + M) gh

V = sqrt(2gh)

V = sqrt(2 x 9.8 x 0.094)

V = 1.36 m/s

d)

v = speed of bullet before collision

V = speed of the combination after collision = 1.36 m/s

m = mass of bullet = 0.010 kg

M = mass of block = 1 kg

Using conservation of momentum

mv = (m + M) V

0.010 v = (0.010 + 1) (1.36)

v = 137.4 m/s

e)

W = change in kinetic energy = (0.5) m v² - (0.5) (m + M) V² = (0.5) (0.010) (137.4)² - 0.93 = 93.5 J