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A neutron having mass of \(1.67 \times 10^{-27} \mathrm{~kg}\) and moving at \(10^{8} \mathrm{~m} / \mathrm{s}\) collides with a deuteron at rest and sticks to it. If the mass of the deuteron is \(3.34 \times 10^{-27} \mathrm{~kg}\) then the speed of the combination is (A) \(3.33 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (B) \(3 \times 10^{5} \mathrm{~m} / \mathrm{s}\) (C) \(33.3 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (D) \(2.98 \times 10^{5} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The short answer is (A) \(3.33 \times 10^{7} \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Determine the initial linear momentums of the neutron and deuteron

To find the initial linear momentum of the neutron, we'll use the formula: Neutron's Initial Linear Momentum = Neutron's mass × Neutron's initial velocity For the deuteron, since it is at rest, its initial linear momentum is zero.
02

Calculate the total initial linear momentum

We need to add the neutron's initial linear momentum to the deuteron's initial linear momentum (which is zero) to find the total initial linear momentum: Total Linear Momentum (Initial) = Neutron's Initial Linear Momentum
03

Determine the final linear momentum of the combined objects

After the collision, the neutron and deuteron stick together and move as one object. We can express their final linear momentum as: Final Linear Momentum = (Neutron's mass + Deuteron's mass) × Final speed
04

Apply the law of conservation of linear momentum

According to the conservation of linear momentum, the total initial linear momentum equals the final linear momentum. Set the expressions from Step 2 and Step 3 equal to each other, then solve for the final speed: Neutron's Initial Linear Momentum = (Neutron's mass + Deuteron's mass) × Final speed
05

Calculate the final speed and find the correct answer

Plug in the known values for the neutron's mass, neutron's initial velocity, and the deuteron's mass into the equation from Step 4. Then, solve for the final speed and find the matching option: \(1.67 \times 10^{-27} kg \times 10^8 m/s = (1.67 \times 10^{-27} kg + 3.34 \times 10^{-27} kg) × Final \: speed\) The final speed can be calculated as: Final speed = \(\frac{1.67 \times 10^{-27} kg \times 10^8 m/s}{(1.67 \times 10^{-27} kg + 3.34 \times 10^{-27} kg)}\) By calculating this equation, we get the final speed and we can find the correct answer among the given options: Final speed ≈ 3 × 10^7 m/s The correct answer is (A) \(3.33 \times 10^{7} \mathrm{~m} / \mathrm{s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic and Inelastic Collisions
Collisions come in two main types: elastic and inelastic. In elastic collisions, both momentum and kinetic energy are conserved. The objects bounce off each other without any lasting deformation or generation of heat. This means the total kinetic energy before and after the collision remains the same. In contrast, inelastic collisions involve the conservation of linear momentum but not kinetic energy. The objects may stick together or deform, converting some kinetic energy into other forms like heat or sound. The collision between the neutron and deuteron in the exercise above is an inelastic collision. This is because the neutron and deuteron stick together after the collision, forming a composite particle. While their combined momentum is conserved, part of the kinetic energy that was initially in the moving neutron is not present in the system after the collision.
Linear Momentum Calculation
Linear momentum is a measure of an object's motion and is calculated by multiplying the object's mass by its velocity: \[ \text{Linear Momentum} = \text{mass} \times \text{velocity} \]In our example, the neutron's linear momentum before the collision is found using its mass and velocity. For the deuteron, since it is initially at rest, its linear momentum is zero. Hence, the total initial linear momentum of the system is equal to the linear momentum of the neutron alone:
  • Neutron's Mass = \(1.67 \times 10^{-27} \mathrm{~kg}\)
  • Neutron's Velocity = \(10^8 \mathrm{~m/s}\)
To find the final speed after the collision, we apply the conservation of momentum, equating the system's initial momentum to its final momentum. Thus:\[ \text{Final Linear Momentum} = (\text{Neutron's Mass} + \text{Deuteron's Mass}) \times \text{Final Speed} \]This allows us to calculate the speed of the new, combined mass after the collision.
Neutron-Deuteron Collision
In a neutron-deuteron collision, a fast-moving neutron impacts a stationary deuteron. A deuteron is essentially a nucleus of deuterium, consisting of one proton and one neutron, and has double the neutron's mass. The collision describes a straightforward scenario where two particles stick together, forming a complex particle right after impact. This type of collision helps in understanding core concepts such as:
  • Conservation Principles - The exercise showcases how linear momentum is conserved in an inelastic collision.
  • Kinetic Energy Transformation - Some of the system's kinetic energy is transformed into other forms during an inelastic collision.
Using known values, we calculated the final speed to demonstrate how mass and speed adjust post-collision in real-life scenarios. Such calculations are crucial in fields like nuclear physics and help researchers design experiments and understand particle interactions.

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Most popular questions from this chapter

Four identical balls are lined in a straight grove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 4 balls from left. What will happen (A) One ball from the right rolls out with a speed \(2 \mathrm{v}\) and the remaining balls will remain at rest. (B) Two balls from the right roll out speed \(\mathrm{v}\) each and the remaining balls will remain stationary. (C) All the four balls in the row will roll out with speed \(\mathrm{v}(\mathrm{v} / 4)\) each and the two colliding balls will come to rest. (D) The colliding balls will come to rest and no ball rolls out from right.

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion : Linear momentum is conserved in both, elastic and inelastic collisions. Reason: Total energy is conserved in all such collisions. (A) 1 (B) 2 (C) 3 (D) 4

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A bomb of mass \(10 \mathrm{~kg}\) explodes into 2 pieces of mass \(4 \mathrm{~kg}\) and \(6 \mathrm{~kg}\). The velocity of mass \(4 \mathrm{~kg}\) is \(1.5 \mathrm{~m} / \mathrm{s}\), the K.E. of mass \(6 \mathrm{~kg}\) is ....... (A) \(3.84 \mathrm{~J}\) (B) \(9.6 \mathrm{~J}\) (C) \(3.00 \mathrm{~J}\) (D) \(2.5 \mathrm{~J}\)

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