Question

A 1000 kilogram empty cart moving with a speed of 6.0 meters per second is about to collide with a stationary loaded cart having a total mass of 5000. kilograms. After the collision, the carts lock and move together. [Assume friction is negligible.]
a) Calculate the speed of the combined carts after the collision. [Show all work, including the equation and substitution with units.]
b) Calculate the kinetic energy of the combined carts after the collision. [Show all work, including the equation and substitution with units.]
c) How does the kinetic energy of the combined carts after the collision compare to the kinetic energy of the carts before the collision?

Answer 1

Answer:

1) v = 1m/s

2) K.E = 3 kJ

3) $K.E_{2} < K.E_{} < K.E_{1}$

Explanation:

From the law of conservation of momentum, p = mv

1) For the given question, $p = p_{1} + p_{2}$ ------- eqn 1

substitute mv for p in eqn 1,

$p_{1} = m_{1} v_{1}$

and $p_{2} = m_{2} v_{2}$

To solve for p, which is the final momentum after collision

$p = m_{1} v_{1} + m_{2} v_{2}$ ------ eqn 2

Parameters

$m_{1}$ = 1000 kg

$m_{2}$ = 5000 kg

$v_{1}$ = 6 m/s

$v_{2}$ = 0 (static object)

substitute for all m and v in eqn 2

p = $(1000 X 6) + (5000 X 0)$

p = 6000 + 0

p = 6000 kgm/s

Recall p = mv

where p = 6000kgm/s and m = $(m_{1} + m_{2})$

v = ?

6000 kgm/s = (1000 + 5000)v

6000 = 6000v

v = 1m/s

2) K.E = $\frac{1}{2} mv^{2}$

Therefore, K.E after collision

K.E = $\frac{1}{2} X 6000 X 1^{2}$

K.E = 0.5 X 6000 X 1 = 3000 J

K.E = 3 kJ

3) Comparing K.Es of the carts, first we write out the K.E of each cart and their combined K.E

(Please ignore the Armstrong in the following equation, I don't know how it appeared from the equation tool)

$K.E_{1} = \frac{1}{2} X 1000 X 6^{2}$ = 18 kJ

$K.E_{2} = \frac{1}{2} X 5000 X 0^{2}$ = 0

Combined K.E = 3 kJ

From the above K.Es,

i) The K.E of the combined cart after collision is lesser than the K.E of the first cart of 1000 kg.

3 kJ < 18 kJ

ii) The K.E of the combined cart is greater than the K.E of the second cart of 5000 kg.

0 kJ < 3 kJ

Conclusion;

$K.E_{2} < K.E_{} < K.E_{1}$

i.e 0 kJ < 3 kJ < 18 kJ

Answer 2

Explanation:

a) m1u1 + m2u2 = v(m1+m2)

1000×6 + 5000×0 = v(1000+5000)

6000 + 0 = 6000v

v = 6000/6000

v = 1 m/s

b) ½ ×(m1+m2)v²

= 0.5×6000×1²

=0.5×6000

=3000J

=3KJ

c) solve for b4 collision and compare

Goodluck