Question

A 6.9 kg block with a speed of 4.8 m/s collides with a 13.8 kg block that has a speed of 3.2 m/s in the same direction. After the collision, the 13.8 kg block is observed to be traveling in the original direction with a speed of 4.0 m/s. (a) What is the velocity of the 6.9 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 13.8 kg block ends up with a speed of 6.4 m/s. What then is the change in the total kinetic energy? (a) Number Units (b) Number Units (c) Number Units

Answer 1

Suppose that the block B is moving with speed v just after the collision.

Now, assume the 2 block system is whole system. As we can see that there is not net external force acting on the system, therefore the linear momentum of the system must conserve.

From linear momentum conservation,

m_Av_Ai + m_Bv_Bi = m_Av_Af + m_Av_Af .........(1)

where

v_Ai = Initial velocity of block A

v_Af = Final velocity of Blcok A after collison

v_Bi = Initial velocity of block B

v_Bf =  Final velocity of block B after collison

Using Equation (1)

6.9×4.8 + 13.8×3.2 = 6.9×v + 13.8×4

⇒ v = 3.2 m/s

(a) Velocity of 6.9 Kg Block will be 3.2 m/s

Change is Kinetic Energy= KE_f - KE_i = ΔKE ......... (2)

KE = ½mv²

ΔKE = [½m_Av_Af²+ ½m_Bv_Bf²] -  [½m_Av_Ai² + ½m_Bv_Bi²]

= - 4.416 J

(b) The change is Total KE is 4.416 J

PART (c)

If the block B ends up with 6.4 m/s. The using momentum conservation equation ...(1). The value of V_Bf will be 6.4

6.9×4.8 + 13.8×3.2 = 6.9×v + 13.8×6.4

v = - 1.6 m/s

This means that block A will move with a speed of 1.6 m/s in the backward direction.

Using equation (2)

ΔKE =  [½m_Av_Af²+ ½m_Bv_Bf²] -  [½m_Av_Ai² + ½m_Bv_Bi²]

ΔKE = 141.312 J

(c) = The change is Kinetic Energy will be 141.312 J.