1. The velocity component normal to the wall (which I will call the x direction) changes from 10.0 sin 60 to -10.0 sin 60. That is a velocity component change of 17.32 m/s. Multiply that by the mass for the momentum change in that direction. Dividing by the contact time gives you the x xomponent of the force.
The velocity component in the y direction does not change, so there is no applied force in that direction.
2. Get the velocity V of the block right after it is hit by applying conservation of energy to the rise of the ballistic pendulum afterwards.
V = sqrt (2gH),
here H = 0.11 m and g = 9.8 m/s^2.
Let v1 and v2 be the velocity of the bullet before and after passing through the block. Apply conservation of linear momentum to that impact
m (v1 - v2) = M V
M is the mass of the block amd m is the mass of the bullet.
Solve for v1
1. A 3.25 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.195 s, what is the average force exerted on the ball by the wall?
x component: N
y component: N
2. A 7.0 g bullet is fired into a 1.5 kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 11 cm. Find the initial speed of the bullet.
2 answers
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials' boat pulls up next to the spy's boat, both boats reach the edge of a 4.8 m waterfall. The spy's speed is 16 m/s and the officials; speed is 26 m/s. How far apart will the two vessels be when they land below the waterfall? The acceleration of gravity is 9.81 m/s . Answer in units of m
1 answer