/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A \(3 \mathrm{~kg}\) ball strike... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 6

A \(3 \mathrm{~kg}\) ball strikes a heavy rigid wall with a speed of \(10 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) with the wall. It gets reflected with the same speed at \(60^{\circ}\) with the wall. If the ball is in contact with the wall for \(0.2 \mathrm{~s}\), the average force exerted on the ball by the wall is (A) \(300 \mathrm{~N}\) (B) Zero (C) \(150 \sqrt{3} \mathrm{~N}\) (D) \(150 \mathrm{~N}\)

Short Answer

Expert verified
The average force exerted on the ball by the wall is (D) 150 N.

Step by step solution

01

Calculate initial and final velocities of the ball

The ball hits the wall at an angle of \(60^\circ\), so we can represent its velocity as a vector. The components of initial velocity will be, \(v_x = 10 \cos{60^\circ} = 5\, m/s\) \(v_y = 10 \sin{60^\circ} = 5\sqrt{3}\, m/s\) As the ball is reflected back at the same angle, the components of final velocity will be, \(v'_x = -5\,m/s\) (since the direction of horizontal component changes after the collision) \(v'_y = 5\sqrt{3}\,m/s\) (the vertical component doesn't change)
02

Find the change in velocity

The change in the horizontal and vertical components of the velocity are as follows: \(\Delta v_x = v'_x - v_x = -5 - 5 = -10\,m/s\) \(\Delta v_y = v'_y - v_y = 0\,m/s\)
03

Calculate the average acceleration

We can calculate the average acceleration using the formula, \(a = \frac{\Delta v}{t}\) The horizontal and vertical components of the average acceleration are: \(a_x = \frac{\Delta v_x}{t} = \frac{-10}{0.2} = -50\,m/s^2\) \(a_y = \frac{\Delta v_y}{t} = \frac{0}{0.2} = 0\,m/s^2\)
04

Calculate the average force

We can calculate the average force exerted on the ball by the wall using Newton's second law, \(F = ma\) The horizontal and vertical components of the average force are: \(F_x = m * a_x = (3\,kg)(-50\,m/s^2) = -150\,N\) \(F_y = m * a_y = (3\,kg)(0\,m/s^2) = 0\,N\) The magnitude of the average force exerted by the wall is: \(F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-150)^2 + 0^2} = 150\,N\) So the correct answer is (D) 150 N.

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Most popular questions from this chapter

Position of two particles are given by \(x_{1}=2 t\) and \(x_{2}=2+3 t\). The velocity of centre of mass at \(t=2 \mathrm{~s}\) is \(2 \mathrm{~m} / \mathrm{s}\). Velocity of centre of mass at \(t=4 \mathrm{~s}\) will be (A) \(2 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1 \mathrm{~m} / \mathrm{s}\) (D) Zero

Two bodies of masses \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are moving with velocities \(10 \mathrm{~m} / \mathrm{s}\) and \(2 \mathrm{~m} / \mathrm{s}\) towards each other. The velocity of their centre of mass is (A) Zero (B) \(1 \mathrm{~m} / \mathrm{s}\) (C) \(2 \mathrm{~m} / \mathrm{s}\) (D) \(4 \mathrm{~m} / \mathrm{s}\)

A metal ball of mass \(2 \mathrm{~kg}\) moving with speed of \(36 \mathrm{~km} / \mathrm{h}\) has a head-on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after collision, both the balls move together, then the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

A body of mass \(m_{1}\) moving with uniform velocity of \(40 \mathrm{~m} / \mathrm{s}\) collides with another mass \(m_{2}\) at rest and then the two together begin to move with uniform velocity of \(30 \mathrm{~m} / \mathrm{s}\). The ratio of their masses \(\frac{m_{1}}{m_{2}}\) is (A) \(0.75\) (B) \(1.33\) (C) \(3.0\) (D) \(4.0\)

A body of mass \(m_{1}\) collides elastically with a stationary body of mass \(m_{2}\) and return with one third speed, then \(\frac{m_{1}}{m_{2}}=\) (A) 1 (B) 2 (C) \(0.5\) (D) \(0.33\)
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