Question

An 8000 kg truck traveling with a velocity of 20. 0 m/s in a direction 30 ° south of east and a 6250 kg car traveling with a velocity of 25. 0 m/s 40 ° north of east collide in a perfectly inelastic collision. Find (a) the final speed and (b) the direction of the combined mass. Also, calculate (c) the change in the kinetic energy of the system.

Answer 1

here,

mass of truck, m1 = 8000 kg
velocity of truck, v1 = 20 m/s
v1x = v1*Cos30 = 20 * 0.866 = 17.32 m/s
V1y = -v1*Sin30 = 20 * 0.5 = -10 m/s

mass of car, m2 = 6250 kg
velocity of car, v2 = 25 m/s
v2x = 25*Cos40 = 25 * 0.766 = 19.15 m/s
v2y = 25*Sin40 = 25 * 0.642 = 16.07 m/s

m1+m2 = 8000 + 6250 = 14250 kg

from conservation of momentum we have :

For X direction:
m1*v1x + m2*v2x = (m1+m2)V'x

solving for final velocity V'x for combined mass :
v'x = (m1*v1x + m2*v2x)/(m1+m2)
v'x = (8000*17.32 + 6250*19.15)/(14250)
v'x = 18.122 m/s

Similarly for y direction :
m1*v1y + m2*v2y = (m1+m2)V'y

solving for final velocity V'x for combined mass :
v'y = (m1*v1y + m2*v2y)/(m1+m2)
v'y = (8000*(-10) + 6250*16.07)/(14250)
v'y = 1.434 m/s

PART A :

Net velocity of combined mass :
Vnet = sqrt(v'x^2 + v'y^2)
Vnet = sqrt((18.122)^2 + (1.434)^2)
Vnet = 18.178 m/s or 18.2 m/s(rounded off)

PART B:

Direction of Combined mass can be found out by using :

TanA = v'y/v'x
A = arcTan(v'y/v'x)
A = arcTan(1.434/18.122)
A = 4.524 degrees North of east with positive x axis

PART C:
KE before collision
KEi = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2
KEi = 0.5 * 8000 * 20^2 + 0.5 * 6250 * 25^2
KEi = 1600000 + 1953125
KEi = 3553125 J

KE after collision :
Kef = 0.5 * (m1 + m2 ) * Vnet^2
KEf = 0.5 * 14250 * 18.178^2
KEf = 2354382.7485 J

Chnage in Energy of system :
D = 3553125 - 2354382.7485
D = 1198742.25 J or 1198.74 kJ