Mitch throws a 130-g lump of clay at a 590-g target,which is at rest on a horizontal surface.After impact,the target,including the attached clay,slides 2.8 m before stopping.If the coefficient of friction is μ = 0.50,find the speed of the clay before impact.
A)6.4 m/s
B)33 m/s
C)29 m/s
D)0.62 m/s
E)34 m/s

Question

Quizplus · Accepted Answer

We can use the conservation of momentum to find the initial speed of the clay.
m1v1 = (m1+m2)v2
where m1 is the mass of the clay, m2 is the mass of the target, v1 is the initial speed of the clay, and v2 is the final velocity of the combined clay and target.
We can also use the work-energy principle to relate the final speed of the combined clay and target to the distance it slid and the coefficient of friction.
W = ΔK
where W is the work done by friction, ΔK is the change in kinetic energy, and can be written as
W = Ff*d = μ*m*g*d
ΔK = (1/2)*(m1+m2)*v2^2
where Ff is the force of friction, d is the distance the target slid, g is the acceleration due to gravity, and μ is the coefficient of friction.
Since the work done by friction removes energy from the system, we can equate the two equations to get
μ*m*g*d = (1/2)*(m1+m2)*v2^2 - (1/2)*m1*v1^2
We can plug in the given values to get
(0.50)*(0.130 kg)*(9.81 m/s^2)*(2.8 m) = (1/2)*(0.130 kg + 0.590 kg)*v2^2 - (1/2)*0.130 kg*v1^2
Solving for v1 gives us
v1 = 29 m/s

Mitch Throws a 130-G Lump of Clay at a 590-G