A bullet with mass m = 5.21 g is moving horizontally with a speed v=367 m/s when it strikes a block of hardened steel with mass M = 14.8 kg(initially at rest). The bullet bounces o the block in a perfectly elastic collision.
(a) What is the speed (m/s) of the block immediately after the collision?
(b) What is the impulse (kg m/s) exerted on the block?
(c) What is the �nal kinetic energy (J) of the block?
(d) How much work (J) did the bullet do on the block?
(e) What was the change in kinetic energy (J) of the bullet?
(f) How much work (J) was done on the bullet?
I will be happy to check your work.
m1 TIMES v1 = m2 TIMES v2
.0521 X 367= 14.8 X ?
.0521 X 367
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14.8
the division should be the answer of wat u seek
ytujytu
To find the answers to the questions, we can use the principles of conservation of momentum and conservation of kinetic energy.
(a) The speed of the block immediately after the collision can be found by using the conservation of momentum. The equation is:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Here,
m1 = mass of the bullet = 5.21 g = 0.00521 kg
v1_initial = initial speed of the bullet = 367 m/s
m2 = mass of the block = 14.8 kg
v2_initial = initial speed of the block (which is at rest) = 0 m/s
v1_final = final speed of the bullet (which is the same as the initial speed) = 367 m/s
Using the conservation of momentum equation, we can solve for v2_final:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
(0.00521 kg) * (367 m/s) + (14.8 kg) * (0 m/s) = (0.00521 kg) * (367 m/s) + (14.8 kg) * v2_final
1.909 kg m/s = 0.019 kg m/s + (14.8 kg) * v2_final
1.890 kg m/s = (14.8 kg) * v2_final
Simplifying further, we get:
v2_final = 1.890 kg m/s / 14.8 kg
v2_final = 0.1277027027027027 m/s
Therefore, the speed of the block immediately after the collision is approximately 0.128 m/s.
(b) The impulse exerted on the block can be calculated by using the equation:
Impulse = m2 * (v2_final - v2_initial)
Here,
m2 = mass of the block = 14.8 kg
v2_final = final speed of the block = 0.128 m/s
v2_initial = initial speed of the block = 0 m/s
Substituting the values, we can find the impulse:
Impulse = (14.8 kg) * (0.128 m/s - 0 m/s)
Impulse = 1.8944 kg m/s
Therefore, the impulse exerted on the block is approximately 1.8944 kg m/s.
(c) The final kinetic energy of the block can be calculated using the equation:
Final kinetic energy = (1/2) * m2 * v2_final^2
Here,
m2 = mass of the block = 14.8 kg
v2_final = final speed of the block = 0.128 m/s
Calculating the final kinetic energy:
Final kinetic energy = (1/2) * (14.8 kg) * (0.128 m/s)^2
Final kinetic energy = 0.11899328 J
Therefore, the final kinetic energy of the block is approximately 0.119 J.
(d) The work done by the bullet on the block can be found using the equation:
Work = Change in kinetic energy of the block
Since the block is initially at rest, the initial kinetic energy is zero.
Therefore,
Work = Final kinetic energy of the block - Initial kinetic energy of the block
Work = 0.119 J - 0 J
Work = 0.119 J
So, the work done by the bullet on the block is 0.119 J.
(e) The change in kinetic energy of the bullet can be calculated using the equation:
Change in kinetic energy = (1/2) * m1 * (v1_final^2 - v1_initial^2)
Here,
m1 = mass of the bullet = 5.21 g = 0.00521 kg
v1_final = final speed of the bullet = 367 m/s
v1_initial = initial speed of the bullet = 367 m/s
Calculating the change in kinetic energy:
Change in kinetic energy = (1/2) * (0.00521 kg) * (367 m/s)^2 - (367 m/s)^2
Change in kinetic energy = 0 J
Therefore, the change in kinetic energy of the bullet is zero.
(f) Since the bullet bounces off the block in a perfectly elastic collision, no work is done on the bullet.
Therefore, the work done on the bullet is zero Joules.