Physics question 1?

Question

Consider the inelastic collision between two objects (A and B). Object A has a mass of 12 kg. and is moving at 8 m/s. Object B has a mass of 4 kg. and is moving at 15 m/s toward object A (Objects A and B are moving towards each other).

(a) What is the velocity and direction of the wreakage after the inelastic collision?



(b)How much energy is dissipated as heat ?

Answer 1

Momentum will be conserved in an inelastic collision

Focus specifically on the equations related to conservation of momentum

P = mv

Right

P1 = 96 KG M/S
P2 = 60 KG M/S

vel = 96 - 60 kg m/s / 16 kg
vel = ? m/s

I think direction (x)

x = arctan (60 / 96)
x = ? degrees

Answer 2

You have to take into account TWO equations
1) conservation of momentum
2) Balance of kinetic energy

12*8-4*15 = 16*V so V =2.25 m/s in the direction of A
Initial kinetic energy = 1/2*12*64 +1/2*4*225 = 834 Joule
Final kinetic energy = 1/2*16*5.0625=40.5Joule
so793.5 joule are transformed into heat and inelastic deformation

Answer 3

Conservation of Momentum
Pi = Pf = (m)(v) + (m)(v) = (m+m)(v)
P(A + B) = (12)(8) +(4)( -15) = 96-60 = 36 kg m/s
P(f) = 36 kg m/s = (16) (v) ===>

A) 2.25 m/s [in the direction of A]

Kinetic Energies Initial and Final
KE(A) = 1/2 (m) (v)^2 = 1/2 * 12 * 64 = 384 kg m^2/s^2
KE(B) = 1/2 (4) (15)^2 = 450 kg m^2/s^2
KE(F) = 1/2 (16) (2.25)^2 = 40.5 kg m^2/s^2
KE(i) = KE(f)
KE(i) = 384+450 = 40.5 + KE(loss)


B) KE(loss) = 793.5 Joules

Answer 4

(a) conservation of momentum:
v=speed of combined body after collision
mava=(ma+mb)v, solve for v give, v=2.25 m/s in direction of object A.

(b) look at kinetic energies before and after.
before: 1/2mava^2+1/2mbvb^2=834 J
after: 1/2(ma+mb)v^2=40.5 J
793.5 J in heat.