A bullet of mass 0.01kg Is fired into a sandbag of mass 0.49kg hanging from a tree. The sandbag, with the bullet embedded into it, swings away at 10m/s.find
a)the momentum after collision
b)the momentum before collision
c)the velocity of the bullet.
ILOVE ALL OF YOU
momentum after=momentum before
momentum before=Mbullet*Vbullet
Momentum after=(.49+.01)10=Mb*Vb
solve for Vbullet
I teaching or sharing my knowledge from which I got from different texbook, internet.
Yes
THE MOMENTUM BEFORE THE COLLISION
102m/s
A boy of mass 40 kg jumps from a wall at 4 m/s. He kneels and stands in 1 second. Calculate the force required to slow it down. How can this force be different if he bends his knee and does not stop in 0.1 second? from given
a)0.5 b)0.5 c)102m/s
e sandbag,with the bullet embedded into it,swings away at 10 meter per second
given m=40Kg, v 4 m/s ,t1=1s ,t2= 0.1s
solution F= ma ,40Kg*4m/s/0.1s
160m/s/0.1s
1600m/s2
Thank you for your positive message! Let's go through the problem step by step to find the answers you're looking for.
a) To find the momentum after the collision, we can use the equation p = m*v, where p is the momentum, m is the mass, and v is the velocity. The momentum is conserved in a collision, so the momentum before and after the collision will be the same. Since the bullet becomes embedded in the sandbag, we can consider the bullet and sandbag as one combined system.
The mass after the collision is the sum of the bullet mass and the sandbag mass, which is 0.01 kg + 0.49 kg = 0.50 kg. The velocity after the collision is given as 10 m/s. Therefore, the momentum after the collision is:
p_after = (mass_after) * (velocity_after)
= 0.50 kg * 10 m/s
= 5 kg·m/s
So, the momentum after the collision is 5 kg·m/s.
b) Since momentum is conserved, the momentum before the collision will also be 5 kg·m/s. Therefore, the momentum before the collision is 5 kg·m/s.
c) To find the velocity of the bullet before the collision, we need to use the concept of conservation of momentum and the information given. The momentum before the collision is equal to the momentum after the collision.
Let's denote the velocity of the bullet before the collision as v_bullet. The mass of the bullet is given as 0.01 kg. The combined mass of the bullet and sandbag after the collision is 0.50 kg. The velocity after the collision is 10 m/s.
Using the equation p_before = p_after, we can write:
(mass_bullet) * (velocity_bullet) = (mass_after) * (velocity_after)
(0.01 kg) * v_bullet = (0.50 kg) * 10 m/s
Solving this equation for v_bullet, we get:
v_bullet = (0.50 kg * 10 m/s) / (0.01 kg)
= 50 m/s
So, the velocity of the bullet before the collision is 50 m/s.
I hope this helps! Let me know if you have any other questions.