Respuesta :
Answer:
Explanation:
m₁ = 3.4 kg m₂ = 6.8 kg
u₁ = 5.7 m/s u₂ = 3.8 m/s
after collision
v₂ = 4.8 m/s
now,
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
3.4 × 5.7 + 6.8 × 3.8 = 3.4 × v₁+ 6.82 × 4.8
v₁ = 3.67 m/s
a) velocity of 3.4 kg block = 3.67 m/s
b) total change in kinetic energy=
[tex]\Delta KE = ({\dfrac{1}{2}mv^2_1 +\dfrac{1}{2}mv^2_2 }) -(\dfrac{1}{2}mu^2_1 +\dfrac{1}{2}mu^2_2 )\\\Delta KE = ({\dfrac{1}{2}\times3.4\times 3.67^2 +\dfrac{1}{2}\times 6.8\times 4.8^2 }) -({\dfrac{1}{2}\times3.4\times 5.7^2 +\dfrac{1}{2}\times 6.8\times 3.8^2 } )\\\Delta KE =-3.09J[/tex]
hence, K E comes out to be -3.09J.
c)
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
3.4 × 5.7 + 6.8 × 3.8 = 3.4 × v₁+ 6.82 × 7.6
v₁ = - 1.94 m/s
change in K E
[tex]\Delta KE = ({\dfrac{1}{2}mv^2_1 +\dfrac{1}{2}mv^2_2 }) -(\dfrac{1}{2}mu^2_1 +\dfrac{1}{2}mu^2_2 )\\\Delta KE = ({\dfrac{1}{2}\times3.4\times (-1.94)^2 +\dfrac{1}{2}\times 6.8\times 4.8^2 }) -({\dfrac{1}{2}\times3.4\times 5.7^2 +\dfrac{1}{2}\times 6.8\times 3.8^2 } )\\\Delta KE = -19.59J[/tex]
hence the change is kinetic energy comes out to be -19.59J
