A pitcher claims he can throw a 0.145-kg baseball with as much momentum as a 3.00-g bullet moving with a speed of 1.50 x 10³ m/s. What must the baseball'sspeed be if the pitcher's claim is valid?Answer:____m/s

Given: Mass of baseball mbaseball=0.145 kg, Mass of bullet mbullet=3 g=3×10-3 kg, Speed of bullet…

Answered: A pitcher claims he can throw a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Concept explainers

Question

A pitcher claims he can throw a 0.145-kg baseball with as much momentum as a 3.00-g bullet moving with a speed of 1.50 x 10³ m/s. What must the baseball's
speed be if the pitcher's claim is valid?
Answer:____
m/s

Expert Solution

Concept

Given: Mass of baseball $m_{b a s e b a l l} = 0.145 k g$ , Mass of bullet $m_{b u l l e t} = 3 g = 3 \times 10^{- 3} k g$ , Speed of bullet $v_{b u l l e t} = 1.5 \times 10^{3} m / s$ .

Required: Speed of baseball $(v_{b a s e b a l l})$ .

Momentum of an object of mass m moving with speed v is given by, $p = m v$ .

According to the question, $p_{b a s e b a l l} = p_{b u l l e t}$ .

$m_{b a s e b a l l} v_{b a s e b a l l} = m_{b u l l e t} v_{b u l l e t}$

$v_{b a s e b a l l} = \frac{m_{b u l l e t} v_{b u l l e t}}{m_{b a s e b a l l}}$

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions