A5.7 g bullet is fired into a 1.5 kg ballistic pendulum. the bullet emerges from the block with a speed of 154 m/s, and the block rises to a maximum height of 11 cm . find the initial speed of the bullet. the acceleration due to gravity is 9.8 m/s 2 . answer in units of m/s.

540.8m/s

Explanation:

From the information giving, the total energy is conserved and the momentum is conserved.

To determine the speed of the ball after the collision, we use the energy conservation rule, I.e

Kinetic Energy of ball after collision = energy to rise to attain height

1/2mv²=mgh

where m,mass of ballistic pendulum=1.5kg,

v=velocity of ballistic pendulum after collision,

g=gravitational acceleration

h=height attain=11cm=0.11m

if we substitute we arrive at

v=√(2gh)

v=√(2*9.8*0.11)

v=1.47m/s.

since we have determine the velocity of the ballistic pendulum after collision, we now use conservation of momentum to determine the initial speed of the bullet.

since

initial momentum=final momentum

mₓ₁vₓ₁+mₐ₁vₐ₁=mₓ₂vₓ₂+mₐ₂vₐ₂

were mₓ₁vₓ₁,mₓ₂vₓ₂ =mass and velocity of ballistic pendulum before and after collision

mₐ₁vₐ₁,mₐ₂vₐ₂=mass and velocity of bullet before and after collision

if we substitute values,we arrive at

(1.5kg*0m/s)+(0.0057kg*vₐ₁)=(1.5kg*1.47m/s)+(0.0057kg*154m/s)

vₐ₁= 3.0828/0.0057

vₐ₁=540.8m/s

answer: good job it's all correct !

explanation:

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s