Question

A 0.032-kg bullet is fired vertically at 230 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?

Answer 1

Answer:

$heigth=83.44m$

Explanation:

Given data

Baseball mass m₁=0.15 kg

initial speed v₁=0

Bullet mass m₂=0.032 kg

final speed v₂=230 m/s

To find

height h=?

Solution

From conservation of momentum we know that

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (0.032kg)(230m/s)+(0.15kg)(0m/s)=(0.15kg+0.032kg)v\\7.36+0=0.182v\\v=7.36/0.182\\v=40.44m/s$

Now from the conservation of mechanical energy

$P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\ (9.8m/s^{2} )h=(1/2)(40.44m/s)^{2}\\ (9.8m/s^{2} )h=(817.7m^{2} /s^{2} )\\h=(817.7m^{2} /s^{2} )/9.8m/s^{2}\\ heigth=83.44m$