A 1 kg mass moving to the right at a speed of 10m/s collides head-on in a perfectly inelastic collision with a stationary 2kg mass. How much KE is lost during the collision?
First I found V_f to be 3.3m/s
then I found KE_i=1/2m_1v_1i^2 +1/2 m2V_2i^2
=1/2(1)(10)^2 +1/2(2)(0)^2
=50J
then KE_f=
=1/2(1)(3.3)^2 +1/2(2)(3.3)^2
=16.335J
then change in KE =16.335-50
=-33.665 J
is this correct? And is the answer negative because this is how much KE was lost?
If they stuck together, it is correct. Perfectly inelastic is without meaning, inelastic would have been enough, however, it should have stated the masses were stuck.
KE lost is 33.7J. I would say KE lost is 33.7J, and forget the negative sign.
Yes, your calculations and reasoning are correct. The initial kinetic energy (KE_i) is calculated by using the formula KE_i = 1/2 m1v1i^2 + 1/2 m2v2i^2, where m1 is the mass of the first object, v1i is the initial velocity of the first object, m2 is the mass of the second object, and v2i is the initial velocity of the second object.
In this case, m1 = 1 kg, v1i = 10 m/s, m2 = 2 kg, and v2i = 0 m/s (since the second object is stationary). Plugging these values into the formula, we get KE_i = 1/2(1)(10)^2 + 1/2(2)(0)^2 = 50 J.
The final kinetic energy (KE_f) can be calculated using the same formula, but this time substitute the final velocities. In this case, the two masses stick together and move with a common final velocity. You correctly calculated the final velocity (V_f) to be 3.3 m/s.
Plugging in m1 = 1 kg, m2 = 2 kg, and V_f = 3.3 m/s, we get KE_f = 1/2(1)(3.3)^2 + 1/2(2)(3.3)^2 = 16.335 J.
The change in kinetic energy (ΔKE) is then given by ΔKE = KE_f - KE_i = 16.335 - 50 = -33.665 J. The negative sign indicates that there was a loss of kinetic energy during the collision. So yes, your answer of -33.665 J is correct, and the negative sign indicates the loss of kinetic energy.