A 4.5 kg block with a speed of 4.2 m/s collides with a 9.0 kg block that has a s

Question

A 4.5 kg block with a speed of 4.2 m/s collides with a 9.0 kg block that has a speed of 2.8 m/s in the same direction. After the collision, the 9.0 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 4.5 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 9.0 kg block ends up with a speed of 5.6 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

m1 = 4.5 kg
v1i = 4.2 m/s
v1f = ?
m2 = 9.0 kg
v2i = 2.8 m/s
v2f = 3.5 m/s

(a)
Using Momentum Conservation,
Initial Momentum = Final Momentum
m1*v1i + m2 * v2i = m1*v1f + m2 * v2f
4.5 * 4.2 + 9.0 * 2.8 = 4.5 * v1f + 9.0 * 3.5
v1f = 2.8 m/s

(b)
K.Ein = 1/2*m1*v1i^2 + 1/2*m2 * v2i^2
K.Ein = 1/2 * 4.5 * 4.2^2 + 1/2 * 9.0 * 2.8^2
K.Ein =  74.97 J

K.Efin = 1/2*m1*v1f^2 + 1/2*m2 * v2f^2
K.Efin =  1/2 * 4.5 * 2.8^2 + 1/2 * 9.0 * 3.5^2
K.Efin = 72.765 J

Change in Total Kinetic Energy = 74.97 - 72.765 J
Change in Total Kinetic Energy = 2.21 J

(c)

Using Momentum Conservation,
Initial Momentum = Final Momentum
m1*v1i + m2 * v2i = m1*v1f + m2 * v2f
4.5 * 4.2 + 9.0 * 2.8 = 4.5 * v1f + 9.0 * 5.6
v1f = -1.4 m/s


K.Efin = 1/2*m1*v1f^2 + 1/2*m2 * v2f^2
K.Efin =  1/2 * 4.5 * 1.4^2 + 1/2 * 9.0 * 5.6^2
K.Efin = 145.53 J

Change in Total Kinetic Energy = 74.97 - 145.53 J
Change in Total Kinetic Energy = 70.56 J

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