The problem involves the conservation of momentum and kinetic energy, as the collision is elastic.

Step 1: Before the collision, the momentum of the system is mv (as the second particle is stationary). After the collision, the momentum of the system is still mv (conservation of momentum).

Step 2: Let's denote the speed of the first particle after the collision as v1 and the speed of the second particle as v2. From the conservation of momentum, we have mv = mv1 + 2mv2.

Step 3: The kinetic energy of the system before the collision is (1/2)mv^2. After the collision, the kinetic energy is (1/2)mv1^2 + (1/2)2mv2^2. From the conservation of kinetic energy, we have (1/2)mv^2 = (1/2)mv1^2 + mv2^2.

Step 4: Solve the equations from step 2 and step 3 to find v1 and v2

Question

The problem involves the conservation of momentum and kinetic energy, as the collision is elastic.

Step 1: Before the collision, the momentum of the system is mv (as the second particle is stationary). After the collision, the momentum of the system is still mv (conservation of momentum).

Step 2: Let's denote the speed of the first particle after the collision as v1 and the speed of the second particle as v2. From the conservation of momentum, we have mv = mv1 + 2mv2.

Step 3: The kinetic energy of the system before the collision is (1/2)mv^2. After the collision, the kinetic energy is (1/2)mv1^2 + (1/2)2mv2^2. From the conservation of kinetic energy, we have (1/2)mv^2 = (1/2)mv1^2 + mv2^2.

Step 4: Solve the equations from step 2 and step 3 to find v1 and v2

Knowee AI · Accepted Answer

The problem involves the conservation of momentum and kinetic energy, as the collision is elastic.

Step 1: Before the collision, the momentum of the system is mv (as the second particle is stationary). After the collision, the momentum of the system is still mv (conservation of momentum).

Step 2: Let's denote the speed of the first particle after the collision as v1 and the speed of the second particle as v2. From the conservation of momentum, we have mv = mv1 + 2mv2.

Step 3: The kinetic energy of the system before the collision is (1/2)mv^2. After the collision, the kinetic energy is (1/2)mv1^2 + (1/2)2mv2^2. From the conservation of kinetic energy, we have (1/2)mv^2 = (1/2)mv1^2 + mv2^2.

Step 4: Solve the equations from step 2 and step 3 to find v1 and v2

A particle of mass m moving with speed v hits elastically another stationary particle of mass 2m inside a smooth horizontal circular tube of radius r. The time after which the second collision will happen?Physics

Question

Solution

Similar Questions

Upgrade your grade with Knowee