Show that the kinetic energy of an object having a mass m and momentum P is give

Question

Show that the kinetic energy of an object having a mass m and momentum P is given by K = P^2/2m. 2. A mass m_t traveling down the x-axis with a speed v undergoes a perfectly inelastic collision with a second mass m_2 that was initially at rest. Show that the total kinetic energy just after the collision. K_r, is [m_1/(m_1 + m_2)]K_i, where K_i is the initial kinetic of the system. 3. If a mass m_a = 2m has initial velocity 4 m/s and mass m_b = m is initially at rest, they undergo elastic collision. Calculate their final velocities after the collision. 4. If a mass m_a = 2m has initial velocity 4 m/s and mass m_b = m has initially velocity -6 m/s, they undergo elastic collision. Calculate the final velocities,

Explanation / Answer

1) KE = 0.5*m*v^2

multiplying with mass m in the numerator and denominator

KE = 0.5*(m*m*v^2)/m = (m*v)^2/(2*m) = p^2/(2m)

--------------------------------------------------

2) in perfectly in elastic collision

m1*v = (m1+m2)*Vf

Vf = (m1*v)/(m1+m2)

KE after the collision Kf= 0.5*(m1+m2)*Vf^2 = 0.5*(m1+m2)*(m1*v)^2/(m1+m2)^2

Kf = 0.5*m1*m1*V^2/(m1+m2) = [m1/(m1+m2)]*0.5*m1*v^2 = m1*Ki/(m1+m2)

---------------------------------------------------

3) final velocity of ma is va = (ma-mb)*u1/(ma+mb)

va = (2m-m)*4/(2m+m) = 4/3 = 1.33 m/s

vb = (2*ma*u1)/(ma+mb) = (2*2m*4)/(m+2m) = 16/3 = 5.33 m/s

-----------------------------------------------------------

4) va = (ma-mb)*u1/(ma+mb) + 2*mb*u2/(m1+mb)

va = (2m-m)*4/(2m+m) - 2*m*6/(2m+m)

va = (4/3)-(4) = -2.67 m/s

vb = 2*ma*u1/(ma+mb) + (mb-ma)*u2/(ma+mb)

vb = 2*2m*4/(2m+m) - (m-2m)*6/(m+2m)

vb = (16/3) + (2) = 7.33 m/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.