/*! 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 46 A body of mass \(m_{1}\) moving ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 46

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) \(4.0\)

Short Answer

Expert verified
The ratio of their masses, $\frac{m_{1}}{m_{2}}$, is found by applying the principle of conservation of linear momentum. The equation is $m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$. Plugging in given values and solving for the ratio, we find that $\frac{m_{1}}{m_{2}} = \frac{3}{1}$, which corresponds to option (C) 3.0.

Step by step solution

01

Identify the principle of conservation of linear momentum

Before the collision, the total momentum of the two masses is the momentum of m1 (since m2 is at rest). After the collision, the total momentum is the momentum of both masses moving together. According to the principle of conservation of linear momentum, the total momentum before the collision must equal the total momentum after the collision.
02

Write the equation for the conservation of linear momentum

Let's denote the initial velocity of m1 as u1, the initial velocity of m2 as u2, and their final velocity as v. Then, the equation for the conservation of linear momentum is: m1u1 + m2u2 = (m1 + m2)v
03

Plug in the given values

Now we can plug the given values into the equation: m1(40) + m2(0) = (m1 + m2)(30)
04

Simplify the equation

Now, let's simplify the equation and solve for the ratio m1/m2: 40m1 = 30m1 + 30m2 10m1 = 30m2 m1/m2 = 30m2/10m1
05

Solve for the ratio

Finally, we can solve for the ratio m1/m2: m1/m2 = 30m2/10m1 m1/m2 = 3/1
06

Match the answer to the options

The ratio m1/m2 = 3/1 which corresponds to the option (C) 3.0. Thus, the ratio of their masses is 3.0.

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Most popular questions from this chapter

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 bomb of mass \(3 m \mathrm{~kg}\) explodes into two pieces of mass \(m \mathrm{~kg}\) and \(2 \mathrm{~m} \mathrm{~kg}\). If the velocity of \(\mathrm{m} \mathrm{kg}\) mass is \(16 \mathrm{~m} / \mathrm{s}\), the total energy released in the explosion is (A) \(192 \mathrm{~m} \mathrm{~J}\) (B) \(96 \mathrm{~m} \mathrm{~J}\) (C) \(384 \mathrm{~m} \mathrm{~J}\) (D) \(768 \mathrm{~m} \mathrm{~J}\)

A moving body collides elastically with another body of equal mass at rest. Then (A) a part of the energy is dissipated as heat. (B) momentum is conserved but \(\mathrm{KE}\) is not conserved. (C) both masses start moving with the same velocity. (D) the moving mass transfers whole of its energy to the mass at rest.

Distance travelled by cart before second collision is (A) \(2 L\) (B) \(2 L\left(\frac{m}{M+m}\right)\) (C) \(2 L\left(\frac{M}{M+m}\right)\) (D) \(L+L \frac{m}{M+m}\)

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
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