   Chapter 6, Problem 41P

Chapter
Section
Textbook Problem

A 0.030-kg bullet is fired vertically at 200 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?

To determine
Height raised by the bullet and baseball after the collision.

Explanation

Given Info: Mass of the ball is M=0.15kg , Mass of the bullet is m=0.030kg , Initial velocity of the bullet is v1i=200m/s , Initial velocity of the baseball is v2i=0 and velocity of ball-bullet combination is V .

The expression for conservation of momentum is,

mv1i+Mv2i=(M+m)V (1)

• m is the mass of the bullet.
• M is the mass of the ball.
• v1i is the initial velocity of the bullet.
• v2i is the initial velocity of the ball.
• V is the velocity of ball-bullet combination.

Rewrite the above equation with zero for v2i .

mv1i=(M+m)VV=mv1iM+m (2)

Apply conservation of energy to ball-bullet system,

(K.E+P.E)after collision =(K.E+P.E)max. height of ball

• K.E is the kinetic energy.
• P.E is the potential energy.

Rewrite the above equation using expression for kinetic and potential energies with zero for both potential energy at the bottom and kinetic energy at the top

12(M+m)V2+0=0+(M+m)ghmax (3)

• g is the acceleration due to gravity

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