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A bullet with mass m = 40 grams traveling at v = 400 m/s strikes a...
A bullet of mass 5 g is fired at 502 m/s horizontally into a block of...
A bullet of mass 5 g is fired at 502 m/s horizontally into a block of wood (2kg) initially at rest suspended on a massless string. The bullet embeds itself in the block of wood and the combination swings up. To what maximum height does the combination swing? Answer in CENTIMETERS with 3 sig figs.
A 10.0 gram bullet traveling at 275 m/s strikes and embeds itself in a 3.490 kg...
A 10.0 gram bullet traveling at 275 m/s strikes and embeds itself in a 3.490 kg block of wood held on a frictionless table by a spring having k= 50.0 kg/sec^2. Calculate the speed of the block immediately after the collision and the compression of the spring in meters.
A 10g bullet traveling at 400 m/s strikes a 10 kg, 1.0m wide door at the...
A 10g bullet traveling at 400 m/s strikes a 10 kg, 1.0m wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?
26) In Figure 8.4, a bullet of mass 0.01 kg moving horizontally strikes a block of...
26) In Figure 8.4, a bullet of mass 0.01 kg moving horizontally strikes a block of wood of which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block? The length of thestringis2 meters. A) 646 m/s B) 366 m/s C)423 m/s D)250 m/s E) 66.7 m/s 667m-
A bullet of mass 0.056 kg traveling horizontally at a speed of 100 m/s embeds itself...
A bullet of mass 0.056 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 1.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? v= m/s (b) Calculate the kinetic energy of the bullet plus the block before the collision: K; = (c) Calculate the kinetic energy of the bullet plus the block after the...
A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is...
A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.68 m long. The block is initially at rest. A bullet with mass 0.0108kg is fired at the block with a horizontal velocity . The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on theend of the cord. When the block has risen a vertical height of 0.775 m, the tension in...
A small wooden block with mass 0.775 kg is suspended from the lower end of a...
A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.74 m long. The block is initially at rest. A bullet with mass 0.0112 kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m ,...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds itself in a block of mass 4 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vf = m/s ) Calculate the total translational kinetic energy before and after the collision. Ktrans,i = Ktrans,f = (c) Compare the two results and explain why there is a...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds itself in a block of mass 5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vr = 42 x m/s (b) Calculate the total translational kinetic energy before and after the collision. Ktrans,i = 374.85 Ktrans,f= (c) Compare the two results and explain why there...
A 10 g bullet traveling at 440 m/s strikes a 13 kg , 1.1-m-wide door at...
A 10 g bullet traveling at 440 m/s strikes a 13 kg , 1.1-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?
A 10g bullet traveling at 400 m/s strikes a 10 kg, 1.0m wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?
26) In Figure 8.4, a bullet of mass 0.01 kg moving horizontally strikes a block of wood of which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block? The length of thestringis2 meters. A) 646 m/s B) 366 m/s C)423 m/s D)250 m/s E) 66.7 m/s 667m-
A bullet of mass 0.056 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 1.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? v= m/s (b) Calculate the kinetic energy of the bullet plus the block before the collision: K; = (c) Calculate the kinetic energy of the bullet plus the block after the...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds itself in a block of mass 4 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vf = m/s ) Calculate the total translational kinetic energy before and after the collision. Ktrans,i = Ktrans,f = (c) Compare the two results and explain why there is a...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds itself in a block of mass 5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vr = 42 x m/s (b) Calculate the total translational kinetic energy before and after the collision. Ktrans,i = 374.85 Ktrans,f= (c) Compare the two results and explain why there...
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