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Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.

User SS Hegde
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1 Answer

6 votes

Answer:

a

The total initial momentum of the two-block system is
p_t = 67.5 \ kg \cdot m/s^2

b

The magnitude of the final velocity of the two-block system
v_f = 1.9014 \ m/s

c

the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is


\Delta KE =- 847.08 \ J

Step-by-step explanation:

From the question we are told that

The mass of first block is
m_1 = 2.50 \ kg

The initial velocity of first block is
u_1 = 27.0 \ m/s

The mass of second block is
m_2 = 33.0\ kg

initial velocity of second block is
u_2 = 0 \ m/s

The magnitude of the of the total initial momentum of the two-block system is mathematically repented as


p_i = (m_1 * u_1 ) + (m_2 * u_2)

substituting values


p_i = (2.50* 27 ) + (33 * 0)


p_t = 67.5 \ kg \cdot m/s^2

According to the law of linear momentum conservation


p_i = p_f

Where
p_f is the total final momentum of the system which is mathematically represented as


p_f = (m_+m_2) * v_f

Where
v_f is the final velocity of the system


p_i = (m_1 +m_2 ) v_f

substituting values


67.5 = (2.50+33 ) v_f


v_f = 1.9014 \ m/s

The change in kinetic energy is mathematically represented as


\Delta KE = KE_f -KE_i

Where
KE_f is the final kinetic energy of the two-body system which is mathematically represented as


KE_f = (1)/(2) (m_1 +m_2) * v_f^2

substituting values


KE_f = (1)/(2) (2.50 +33) * (1.9014)^2


KE_f =64.17 J

While
KE_i is the initial kinetic energy of the two-body system


KE_i = (1)/(2) * m_1 * u_1^2

substituting values


KE_i = (1)/(2) * 2.5 * 27^2


KE_i = 911.25 \ J

So


\Delta KE = 64.17 -911.25


\Delta KE =- 847.08 \ J

User Jonathan Beebe
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