Question

A bullet of mass 148 g and a speed v impacts on a pendulum of mass 6.18 kg and length 200 cm. The bullet embeds itself in the pendulum bob. The pendulum then travels in a complete and perfect circle.

(a) If the pendulum is suspended by a light, stiff rod, what is the minimum speed of the bullet?


(b) If the pendulum is suspended by a string, what is the minimum speed of the bullet?

Answer 1

here,

mass of bullet , m1 = 148 g = 0.148 kg

mass of pendulam , m2 = 6.18 kg

length , l = 200 cm = 2 m

a)

when it is suspended by a light stiff rod

for complete circle it just achive the top point of circle

so,

the speed after the collison with bullet be v

using conservation of energy

0.5 * (m1 + m2) * v^2 = (m1 + m2) * g * (2L)

0.5 * v^2 = 9.81 * ( 2 * 2)

v = 8.86 m/s

let the speed of bullet before the collison be u

using conservation of momentum for the collison

m1 * u = (m1 + m2) * v

0.148 * u = (0.148 + 6.18) * 8.86

solving for u

u = 378.8 m/s

the initial speed of bullet is 378.8 m/s

b)

when it is suspended by a light stiff string

for complete circle it just achive the top point of circle and centripital force must be equal to the gravitational force

let the speed at the top be v1

equating the forces,

(m1 + m2) * v1^2 /l = (m1 + m2) * g

v1 = sqrt(9.81 * 2) = 4.43 m/s

so,

the speed after the collison with bullet be v

using conservation of energy

0.5 * (m1 + m2) * v^2 = (m1 + m2) * g * (2L) + 0.5 * (m1 + m2 ) * v1^2

0.5 * v^2 = 9.81 * ( 2 * 2) + 0.5 * 4.43^2

v = 9.91 m/s

let the speed of bullet before the collison be u

using conservation of momentum for the collison

m1 * u = (m1 + m2) * v

0.148 * u = (0.148 + 6.18) * 9.91

solving for u

u = 423.7 m/s

the initial speed of bullet is 423.7 m/s