Please answer all parts asap!!! y, a small piece of putty of mass m₁ isInfired with velocity v₁¡ at a stationary wooden block of mass m₂ resting on a horizontal table. After ahead-on collision, the putty sticks to the wooden block and the two slide along the surface of thetable together as a single mass.(A) What equations from conservation are useful for determining V₁f and v2f. Simplify equationsand solve for vf in terms of v₁; and the two masses m₁ and m2.(B) After the collision the two masses slide along the table a distance x before being slowed byfriction and coming to a stop. The coefficient of kinetic friction uk. Solve for the stopping distancex in terms of the masses m₁ and m2, the initial fired velocity of the putty v₁i, gravity g, and thecoefficient of kinetic friction μk.(C) Before the experiment the mass of the putty is determined to be 105 grams, the mass of theblock is 10.5 kg and the putty is fired with an initial speed of 25 m/s. The coefficient of frictionbetween the block and the table is determined to be 0.15. Determine the expected stoppingdistance x.

Name: Please answer all parts asap!!! y, a small piece of putty of mass m₁ i
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: Please answer all parts asap!!! y, a small piece of putty of mass m₁ isInfired with velocity v₁¡ at a stationary wooden block of mass m₂ resting on a horizontal table. After ahead-on collision, the putty sticks to the wooden block and the two slide a

Given:- A small piece of putty of mass m1 moving with the velocity v1i and the mass of the wooden block is m2 and it is on rest initially. after the collision, the putty sticks to the wooden block, and the two slide along the surface of the table. m1= 105 gran m2 = 10.5 kg The initial speed of putty is v1i = 25 m/s The coeffcient of friction between the block and table is μ= 0.15 Find:- The expected stopping distance? Solution:- A) We know that the law of conservation of momentum, we know that the collision of the putty and block is an inelastic collision. Let the final velocity of the both this vf. So now m1v1i+m2×0= m1+m2vfvf = m1v1i(m1+m2)------1) This is an expression of the final velocity of the block and putty. B) We know that the friction force that acts here on a block and putty combined mass is given as fr =μ m1+m2g Now work done by the friction is given as Wfr = frx due to this friction work the block comes to rest, So We know that the law conservation of energy is given as 12m1+m2vf2 = Wfr12m1+m2vf2 = frx12m1+m2vf2 =μm1+m2gxvf22=μgxx= vf22μg---------2) This is the expression of the expected stopping distance.C) Put the given value in equation 1) and get as vf = 0.105×250.105+10.5vf = 0.2475≈0.25vf =0.25 m/s This is the final velocity after the collision. Now put the given value in the eq2) x = 0.2522×0.15×9.8x = 0.02125≈0.021x = 0.021 m So this is the expected stopping distance.C) The final velocity after the collision is 0.25 m/s. The Expected stopping distance is 0.021 m.