Question

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italian ham. The slices of ham are weighed ona plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m. The slices of ham are dropped on the plate all at the sametime from a height of 0.250 m. They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume thatthe collision time is extremely small.

What is the amplitude of oscillation A of the scale after the slices of ham land on the plate?
Express your answer numerically in meters and take free-fall acceleration to be g = 9.80 m/s^2.

Answer 1

General guidance

Concepts and reason

The concept of the conservation of momentum in inelastic collision, potential energy of the spring-mass system and conservation of energy principle.

Initially, use the conservation of energy principle to find the speed of ham before collision. Find the final speed of the spring-mass system by using the conservation of momentum principle.

Later, use the energy conservation principle and equate initial kinetic energy of spring-mass system equals to the potential energy of spring at maximum amplitude.

Finally, use the expression of period of oscillation for spring-mass system.

Fundamentals

The conservation of momentum states that the momentum of the system remains conserved in the inelastic collision if there are no external forces acting on the system.

For a collision occurring between two masses ${m_1}$and ${m_2}$in an isolated system, the total momentum of the two masses before the collision is equal to the total momentum after the collision.

The equation of the conservation of momentum is given as follows:

${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Here, ${m_1}$is the mass object 1with initial velocities ${u_1}$and ${m_2}$is the mass of object 2 moving with initial velocity ${u_2}$. ${v_1}$and ${v_2}$are the final velocities of both the objects after the collision.

The conservation of energy states that potential energy at height h must equal to the kinetic energy at the bottom. The expression is given as follows:

$\\begin{array}{c}\\\\\\frac{1}{2}m{v^2} = mgh\\\\\\\\v = \\sqrt {2gh} \\\\\\end{array}$

The potential energy of spring system at maximum amplitude is given as follows:

$PE = \\frac{1}{2}k{A^2}$

Here, A is the amplitude of oscillation.

The expression of the period of the oscillation of the spring mass system is given as follows:

$T = 2\\pi \\sqrt {\\frac{{m + M}}{k}}$

Here, m+M is the net mass of the system and k is the spring constant.

Step-by-step

Step 1 of 2

(a)

The conservation of energy states that potential energy at height h must equal to the kinetic energy at the bottom. The expression is given as follows:

$\\begin{array}{c}\\\\\\frac{1}{2}m{v^2} = mgh\\\\\\\\v = \\sqrt {2gh} \\\\\\end{array}$

Substitute $9.8{\\rm{ m/}}{{\\rm{s}}^2}$for g and 0.25 m for h in the expression $v = \\sqrt {2gh}$.

$\\begin{array}{c}\\\\v = \\sqrt {2\\left( {9.8{\\rm{ m/}}{{\\rm{s}}^2}} \\right)\\left( {0.25{\\rm{ m}}} \\right)} \\\\\\\\ = 2.21{\\rm{ m/s}}\\\\\\end{array}$

Using the momentum conservation, find the final speed of the spring-mass system.

$\\begin{array}{c}\\\\mv = \\left( {m + M} \\right){v_f}\\\\\\\\{v_f} = \\frac{{mv}}{{m + M}}\\\\\\end{array}$

Here, m is the mass of Italian ham, M is the mass of plate and v is the speed of ham before collision.

Substitute 0.3 kg for m, 0.4 kg for M and 2.21 m/s for v in the above expression.

$\\begin{array}{c}\\\\{v_f} = \\frac{{\\left( {0.3{\\rm{ kg}}} \\right)\\left( {2.21{\\rm{ m/s}}} \\right)}}{{0.3{\\rm{ kg}} + 0.4{\\rm{ kg}}}}\\\\\\\\ = 0.947{\\rm{ m/s}}\\\\\\end{array}$

Again, using the conservation of energy, the kinetic energy of the system is equal to the potential energy of spring at maximum amplitude.

$\\begin{array}{c}\\\\KE = PE\\\\\\\\\\frac{1}{2}\\left( {m + M} \\right){v_f}^2 = \\frac{1}{2}k{A^2}\\\\\\\\A = \\sqrt {\\left( {\\frac{{m + M}}{k}} \\right)} {v_f}\\\\\\end{array}$

Substitute 0.3 kg for m, 0.4 kg for M, 200 N/m for k and 0.947 m/s for ${v_f}$ in the above expression$A = \\sqrt {\\left( {\\frac{{m + M}}{k}} \\right)} v$.

$\\begin{array}{c}\\\\A = \\sqrt {\\left( {\\frac{{0.3{\\rm{ kg}} + 0.4{\\rm{ kg}}}}{{200{\\rm{ N/m}}}}} \\right)} \\left( {{\\rm{0}}{\\rm{.947 m/s}}} \\right)\\\\\\\\ = 0.056{\\rm{ m}}\\\\\\end{array}$

Part a

The amplitude of oscillation of the scale is $5.6 \\times {10^{ - 2}}\\,{\\rm{m}}$.

Explanation | Common mistakes | Hint for next step

At the top, the maximum potential energy of the ham is equal to the kinetic energy at the bottom. Hence, the speed of ham before the collision is evaluated using the energy conservation principle.

In the inelastic collision, momentum is conserved but the kinetic energy is not conserved, because of this conservation momentum is used. At last, when the slice of ham land on the plate, then the kinetic energy of system is equals to the maximum potential energy of spring.

Step 2 of 2

(b)

Substitute 0.3 kg for m, 0.4 kg for M and 200 N/m for k in the expression of period of oscillation $T = 2\\pi \\sqrt {\\frac{{m + M}}{k}}$.

$\\begin{array}{c}\\\\T = 2\\pi \\sqrt {\\frac{{0.3{\\rm{ kg}} + 0.4{\\rm{ kg}}}}{{200{\\rm{ N/m}}}}} \\\\\\\\ = 0.371{\\rm{ s}}\\\\\\end{array}$

Part b

The period of oscillation of the scale is 0.371 s.

Explanation

The time taken to complete one oscillation is known as period of oscillation. It is measured in seconds.

Answer

Part a

The amplitude of oscillation of the scale is $5.6 \\times {10^{ - 2}}\\,{\\rm{m}}$.

Part b

The period of oscillation of the scale is 0.371 s.

Answer only

Part a

The amplitude of oscillation of the scale is $5.6 \\times {10^{ - 2}}\\,{\\rm{m}}$.

Part b

The period of oscillation of the scale is 0.371 s.

Answer 2

5.80*10^-2 m

Answer 3

speed of ham before colission is

V1= sqrt(2*g*H) = sqrt(2*9.81*0.25) =2.215 m/s

 

The collision is inelastic. Liniar momentum is the same before and after collision.

m*V1 =(m+M)*V2

V2 = mV1/(m+M) =0.3*2.215/(0.3+0.4) =0.949 m/s

 

Now the initial kinetic energy of the masses+ sping = potential energy at maximum apmlitude

(m+M)*v^2/2 = k*A^2/2

A = v*sqrt[(m+M)/k) =0.0561 m= 5.61 cm

 

Period of oscillation is

T = 2*pi*sqrt((m+M)/k) =0.3717 seconds

 

Answer 4

answer: 0.02267 m