A steel ball of mass 0.50 kg, moving with a velocity 2.0 m/s, strikes a second ball of mass 0.30 kg, initially at rest. The collision is a glancing one, causing the first ball to be deflected by an angle of 30°, with a speed of 1.50 m/s. Determine the velocity of the second ball after the collision, giving both its speed and direction 16. A steel ball of mass 0.50 kg, moving with a velocity2.0 m/s, strikes a second ball of mass 0.30 kg, initially atrest. The collision is a glancing one, causing the first ball to bedeflected by an angle of 30°, with a speed of 1.50 m/s.Determine the velocity of the second ball after the collision,giving both its speed and directionBefore the collisionofAfter the collisionbᎾ

Given Data:Mass of the first ball Initial velocity of the first ball (Before collision) Mass of the…

Answered: A steel ball of mass 0.50 kg, moving…

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

Question

A steel ball of mass 0.50 kg, moving with a velocity 2.0 m/s, strikes a second ball of mass 0.30 kg, initially at rest. The collision is a glancing one, causing the first ball to be deflected by an angle of 30°, with a speed of 1.50 m/s. Determine the velocity of the second ball after the collision, giving both its speed and direction

16. A steel ball of mass 0.50 kg, moving with a velocity
2.0 m/s, strikes a second ball of mass 0.30 kg, initially at
rest. The collision is a glancing one, causing the first ball to be
deflected by an angle of 30°, with a speed of 1.50 m/s.
Determine the velocity of the second ball after the collision,
giving both its speed and direction
Before the collision
of
After the collision
b
Ꮎ

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 14 images

SEE SOLUTION Check out a sample Q&A here

Follow-up Questions

Read through expert solutions to related follow-up questions below.

Follow-up Question

The conservation of energy equation used in the answer was:
Conservation of momentum: m1v1 = m1vf1 + m2vf2

I believe the conservation of energy equation is:
m1v1 + m2v2 = m1vf1 + m2vf2

Therefore the answer should be 1.17 m/s.

Am I correct? Can someone check my work please

Solution

by Bartleby Expert

SEE SOLUTION

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