A 6.2 kg block with a speed of 4.2 m/s collides with a 12.4 kg block that has a

Question

A 6.2 kg block with a speed of 4.2 m/s collides with a 12.4 kg block that has a speed of 2.8 m/s in the same direction. After the collision, the 12.4 kg block is observed to be traveling in the original direction with a speed of 3.5 m/s. (a) What is the velocity of the 6.2 kg block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 12.4 kg block ends up with a speed of 5.6 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

here,

m1 = 6.2 kg , u1 = 4.2 m/s

m2 = 12.4 kg , u2 = 2.8 m/s

final speed of 2 , v2 = 3.5 m/s

a)

let the final speed of 1 be v1

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

6.2 * 4.2 + 12.4 * 2.8 = 6.2 * v1 + 12.4 * 3.5

solving for v1

v1 = 2.8 m/s

b)

the change in total kinetic energy , KE = 0.5 * m1 * (v1^2 - u1^2) + 0.5 * m2 * ( v2^2 - u2^2)

KE = 0.5 * 6.2 * ( 2.8^2 - 4.2^2) + 0.5 * 12.4 * ( 3.5^2 - 2.8^2)

KE = -3.04 J

c)

v2 = 5.6 m/s

let the final speed of 1 be v1

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

6.2 * 4.2 + 12.4 * 2.8 = 6.2 * v1 + 12.4 * 5.6

solving for v1

v1 = - 1.4 m/s

d)

the change in total kinetic energy , KE = 0.5 * m1 * (v1^2 - u1^2) + 0.5 * m2 * ( v2^2 - u2^2)

KE = 0.5 * 6.2 * ( 1.4^2 - 4.2^2) + 0.5 * 12.4 * ( 5.6^2 - 2.8^2)

KE = 97.2 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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