Classical Mechanics - Block Friction Problem

[TomAlso]

Homework Statement

A block of mass $$M$$ is placed on a horizontal plane with kinetic friction coefficient equal to $$\mu$$. A particle with mass $$m$$ and velocity $$\vec{v_0}$$ hits the block in a completely inelastic collision (i.e. the two bodies get stuck together). Knowing that the angle between $$\vec{v_0}$$ and the horizontal plane is $$\theta$$, calculate the final velocity of the block and the impulse between block and particle in the collision. (See attached picture).

Homework Equations

Impulse-momentum theorem

The Attempt at a Solution

I really do not know where to start from, but if I had to guess I would use conservation of momentum on the horizontal and vertical axis to begin:

on the $$x$$ axis: $$m v_0 \cos \theta = (M+m)v_f + J_F$$

on the $$y$$ axis: $$m v_0 \sin \theta = J_N$$

where $$J_F \mbox{ and } J_N$$ are the impulses of the friction force and normal reaction respectively. Then I would need a third equation, maybe with energy conservation. Someone has a different approach or would like to make any suggestion?? Thanks in advance for your help.

 

[lippyka]

The equations you have written are correct. Now, to get the third equation, use the fact that the collision is inelastic. This means that the total kinetic energy before and after the collision must be equal.Therefore, we have:mv_0^2/2 + MV^2/2 = (M + m)Vf^2/2Where V is the initial velocity of the block.Now you have three equations with three unknowns (Vf, Jf, and Jn). Solve these equations simultaneously to get the answer.

 

[nlightner]

I would approach this problem by first defining the variables and parameters involved. Then, I would use the known equations and principles of classical mechanics to solve for the unknowns.

In this problem, we have a block with mass M and a particle with mass m colliding on a horizontal plane with kinetic friction coefficient \mu. The block has an initial velocity of zero and the particle has an initial velocity \vec{v_0} at an angle \theta with the horizontal plane. After the collision, the two bodies become stuck together and we are asked to find the final velocity of the block and the impulse between the block and particle.

To begin, we can use the impulse-momentum theorem to relate the change in momentum of the block and particle to the impulses acting on them during the collision. We can also use the conservation of momentum to relate the initial and final momenta of the system.

On the x-axis, we can write:

m v_0 \cos \theta = (M+m)v_f + J_F

Where v_f is the final velocity of the block and particle together and J_F is the impulse of the friction force.

On the y-axis, we can write:

m v_0 \sin \theta = J_N

Where J_N is the impulse of the normal reaction force.

To find the final velocity v_f, we can use the conservation of momentum equation:

m v_0 = (M+m)v_f

Solving for v_f, we get:

v_f = \frac{m v_0}{M+m}

To find the impulse of the friction force, we can use the fact that the block and particle are stuck together after the collision, so their relative motion is zero. This means that the impulse of the friction force must be equal in magnitude to the change in momentum of the system, which we can calculate using the conservation of momentum equation:

J_F = (M+m)v_f - m v_0

Substituting in the value of v_f, we get:

J_F = \frac{m v_0}{M+m} - m v_0

To find the impulse of the normal reaction force, we can use the equation for the impulse on the y-axis and solve for J_N:

J_N = m v_0 \sin \theta

Finally, to find the total impulse between the block and particle, we can use the Pythagorean theorem:

J = \sqrt{