Question

A 4.5-kg block moving at 2.0 m/s west on a frictionless surface collides totally inelastically with a second 1.0-kg block traveling east at 2.0m/s.

a) Determine the final velocity of the blocks.

b)Determine the kinetic energy of the first block before the collision.

c)Determine the kinetic energy of the second block before the collision.

d)Determine the kinetic energy of the first block after the collision.

e)Determine the kinetic energy of the second block after the collision.

Answer 1

Mass of the first block is

Mass of the second block is

Initial velocity of the first block

V1 = 2.0 m/s

Initial velocity of the second block

We have taken the velocity in the west direction to be positive and the velocity in the east direction to be negative.

(a)

Initial momentum of the system of two blocks is

Pi= m1v1 + m2uz

In totally inelastic collision, the two blocks stick together after the collision, and move together. with the same velocity. Let v be the final velocity of the two blocks. The final momentum of the blocks is

$P_f=m_1v+m_2v$

$P_f=(m_1+m_2)v$

Applying conservation of momentum

$P_f=P_i$

$v=\frac{7.0\ kgm/s}{5.5\ kg}$

$\boxed{v=1.27\ kgm/s}$

(b)

Kinetic energy of the first block before the collision is

$K_{1i}=\frac12m_1v_{1i}^2$

$K_{1i}=\frac12\times(4.5\ kg)\times(2.0\ m/s)^2$

$\boxed{K_{1i}=9.0\ J}$

(c)

Kinetic energy of the second block before the collision is

$K_{2i}=\frac12m_2v_{2i}^2$

$K_{2i}=\frac12\times(1.0\ kg)\times(-2.0\ m/s)^2$

$\boxed{K_{2i}=2.0\ J}$

(d)

Kinetic energy of the first block after the collision is

$K_{1f}=\frac12m_1v^2$

$K_{1f}=\frac12\times(4.5\ kg)\times(1.27\ m/s)^2$

$\boxed{K_{1f}=3.63\ J}$

(e)

Kinetic energy of the second block after the collision is

$K_{2f}=\frac12m_2v^2$

$K_{2f}=\frac12\times(1.0\ kg)\times(1.27\ m/s)^2$

$\boxed{K_{2f}=0.806\ J}$